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YouTube transcript (tsPv-ffN-0M) — Transcript

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0:09Let's go ahead and look at a couple of
0:10examples. So, a report from 6 years ago
0:13indicated that the average gross salary
0:16for a business analyst was $69,873.
0:21Now, since this survey is now outdated,
0:24the Bureau of Labor Statistics wishes to
0:26test this figure against current
0:29salaries to see if the current salaries
0:32are statistically different from the old
0:34ones. So, the
0:37$69,873 will be our presumed our assumed
0:41population mean that we're going to test
0:44against. Now, based on other studies,
0:46we're going to assume a sigma of
0:52$13,985. Now, for this study, the BLS
0:55will take a sample of 112 current
0:59salaries. So, we have all the parts we
1:02need to set up our hypothesis. We have
1:05the hypothesized population mean. We
1:08have our sigma. We have our sample size.
1:11And of course once we collect our data,
1:13we'll have our sample
1:17mean. Now step one is always establish
1:20our hypothesis. Now we went ahead and
1:22did the other step one which is
1:24formulate a good problem. We did that in
1:25the previous slide. So actually in the
1:27numbers we have step one establish the
1:30hypothesis. So remember our null
1:33hypothesis is that the current salary
1:36mean is the same as the previous one 6
1:40years ago. So mu is equal to
1:4669,873. Now our alternative is the
1:48opposite of that and that is that the
1:50current mean salary for business
1:53analysts is not $69,873.
1:59Now, step two, determine the appropriate
2:01statistical test and sampling
2:03distribution. Now, this will be a
2:05two-tailed test because remember,
2:06salaries could be higher or lower
2:09because we're still in the middle of a
2:12global recession. So, it's very possible
2:15that salaries could have gone down. Now,
2:17they could have gone up as well. We
2:18don't know. So, we're just testing
2:20whether or not it's equal to. We don't
2:22know on which side it may have gone if
2:25it's not equal to. Now, since sigma is
2:28known, we will be using the Z
2:30distribution as we will in all the
2:32examples in this video. So, we're going
2:34to go ahead and use the formula we had
2:36two slides
2:40ago. Now, step three, we're going to
2:42specify the type one error rate or the
2:45significance level. And again, this is
2:47up to us. So, I'm going to choose for
2:49this one the middle ground and say an
2:51alpha of 0.05.
2:54So I am okay with the possibility of
2:58making a type one error 5% of the time.
3:03Now step four, we're going to state our
3:06decision rule. Now remember based on the
3:10curve, if our ZV valueue is above
3:141.96, it'll be in our top rejection
3:17region. Therefore, we'll reject the null
3:20hypothesis. If our zstistic is less than
3:24negative 1.96 then again we will reject
3:27the null hypothesis because that will be
3:29in our lower rejection region. Now of
3:33course step five we will gather the
3:35data. Now in this case we went ahead and
3:38gathered our data. So our sample size
3:40was 112 and our sample mean was 70
3:4872,180. So, we know that it's higher,
3:52but the question is, is it high enough
3:55to be statistically
4:01significant? Let's go ahead and
4:03calculate our test statistics. So, our
4:05mean salary for the current salaries is
4:11$72,180. Our
4:13hypothesized mean was
4:16$69,873. Now our sigma was given to us
4:19at 13985 and of course our sample size
4:22is 112. So again we have our formula
4:25down here at the bottom. So all we do is
4:28we go ahead and insert those numbers
4:31into our Z formula. So 72180 which is
4:36our sample mean minus
4:3969,873 which is our hypothesized mean
4:42divided by sigma / the<unk> of n. And
4:47that comes up with a value of Z that is
4:50equal to
4:531.75. Now we have a decision to
4:57make. So remember our hypothesis were mu
5:01is
5:0369,873 or the alternative was mu is not
5:0869,873. Now we know it's not exactly
5:1269,873 because we found that it was like
5:1471,000 something. But the question is,
5:17is it high enough to say that it is
5:21statistically
5:23different? So our Z was
5:271.75. And look where that falls in our
5:30non-rejection region. It's right there
5:33to the left of our critical value. Now
5:37since the test statistic is inside the
5:39non-rejection region and not beyond the
5:42critical value, we fail to reject the
5:46null hypothesis that the old and the
5:49current salaries are statistically
5:52different. So we therefore fail to
5:57reject the null hypothesis. So we assume
6:01that our assumption holds up. Remember,
6:04we're not saying that our null is quote
6:07true. All we're saying is that we could
6:10not reject it based on the statistical
6:13analysis we did. So, is the salary
6:17higher based on our sample? Yes. But can
6:21we say it is statistically different?
6:24No. And remember why is that? That's
6:28because of the idea of sampling error.
6:32Remember, this was just one sample. We
6:35could have taken many samples. And those
6:38other samples might be right smack in
6:40the middle of the non-rejection region.
6:42We might have a sample that's lower in
6:45the non-rejection region. They could be
6:47anywhere in that blue area. We just
6:50happened to get one here. Now remember
6:54what we're saying is that 5% of the time
6:57we expect to get a sample mean that's
7:00either in the upper rejection region or
7:02in the lower rejection region. But this
7:05one sample just happens to be located
7:08right here right at the upper edge of
7:11the non-rejection region. So we would
7:13conclude that the old salaries and the
7:16current salary are not
7:19statistically different.
7:24So, example two, Starbucks customer
7:26satisfaction. So, Starbucks is
7:28interested in assessing customer
7:31satisfaction in the Canadian city of
7:33Toronto, Ontario. To conduct the study,
7:36Starbucks asks 225 customers in the city
7:41compared to other coffee houses in
7:43Toronto. Would you say the customer
7:45service at Starbucks is much better than
7:47average, which is a score of five?
7:49better than average, which is a score of
7:51four. Average, a score of three. Worse
7:54than average, a score of two. Or much
7:57worse than average, a score of one. And
8:01this is commonly known as a Lykert
8:03scale. So 54321 in descending order. Now
8:08based on the data we collected, the mean
8:10rating was determined to be 3.25.
8:14And based on previous studies done by
8:16the company, it is assumed that sigma is
8:221.5. So let's go ahead and establish our
8:25hypothesis. So our null hypothesis is
8:29that the average customer rating is less
8:32than or equal to three because remember
8:34three is average in our lacquered scale.
8:37Three is average. And then our
8:39alternative hypothesis is that the
8:42satisfaction level is higher than three.
8:46So we're going to assume that it's three
8:49or less and then we'll either reject or
8:52fail to reject that and then we will go
8:55on to our alternative because remember
8:57the equal sign the equality portion is
9:00always in the
9:01null. So determine the appropriate
9:03statistical test and sampling
9:05distribution. Now, this will be a
9:07onetailed test. Starbucks is interested
9:11in a better than average customer
9:14service rating. So, you can see that in
9:17our alternative hypothesis, we're
9:20interested if the the average customer
9:22rating is higher than three. So, because
9:26of the greater than and less than or
9:28equal to than, this will be a one-
9:29tailed test. Now since sigma is known,
9:32we will again use the Z distribution as
9:34we did
9:37before. Now specify the type one error
9:41rate. So our significance level. Now for
9:44this one, I'm going to choose a 01.
9:47Again, just to show you some conceptual
9:49information, I'm going to show you a
9:50different one or use a different one.
9:52Then we'll state the decision rule. So
9:55if our ZV valueue is greater than
10:002.33, we will reject the null
10:03hypothesis. So this is our upper tailed
10:05test. Remember with an alpha of 01, we
10:08have 99% there in the non-rejection
10:10region and we have the 1% all in the
10:14upper rejection region. So our Z, our
10:17critical value is 2.33, which is right
10:21there. Of course, where does that come
10:22from? that comes from our Z table. So
10:26you just have to look up that critical
10:29value in the Z table. Of course, step
10:33five will gather our data. Now, we
10:34already did that. So our sample size
10:36again was 225 and our sample mean was
10:453.25. So let's go ahead and calculate
10:47our test statistic. So there are four
10:50inputs. our sample mean, our
10:52hypothesized mean, our sigma, and our
10:55sample size. And again, the same
10:57formula. So, we'll go ahead and
10:59substitute all that information into our
11:02equation, and we come up with a ZV
11:05value, a Z test value of
11:092.5. So, let's go ahead and put that on
11:11our
11:14curve. So, here is our distribution, and
11:17you can see that we have our
11:18non-rejection region in the middle.
11:20That's 99% or 0.99, that's our
11:22probability in the middle. And of
11:24course, we have our 1% there on the
11:26ends. So remember what we're saying here
11:29is that we expect 99% of our sample
11:32means to be in this blue region and then
11:351% to be in the rejection region in the
11:39uh brown color there. So here our
11:41hypothesis, our null is that mu is less
11:44than or equal to 3 and the alternative
11:46is that mu is greater than three.
11:49So our Zcritical value is 2.33 which is
11:53right there on our curve. Now our Z is
11:582.5. So where is it at? It is in the
12:03rejection region. Now since the test
12:07statistic is inside the rejection region
12:10and beyond the critical value, we reject
12:14the null hypothesis that customer
12:16satisfaction is at or below average. So
12:20therefore we have to reject the null
12:23hypothesis and accept the alternative
12:27hypothesis based on this Z value.
12:35Now, what is the mean customer
12:37satisfaction value at the critical value
12:41of
12:422.33? Because remember, up until now,
12:44we've been finding zcores. But what is
12:47the actual customer sentiment, the
12:50average customer satisfaction at that
12:54critical
12:55value? Well, remember this is the
12:57formula we used before. So, it's
12:59everything we had in our equation
13:00before. Now what we can do is we can
13:04just sort of put in the Zcritical value
13:07we're looking at. So 2.33 just goes in
13:10where Z was. Now we're solving for Xbar.
13:16So we want to find the sample that would
13:19fall right on the sample mean that would
13:21fall right on that Z critical value. So
13:24again, this is just where our wonderful
13:27basic algebra that we learned many years
13:29ago comes into play.
13:32So we'll go ahead and solve our
13:33denominator. That ends up being 0.1.
13:36Then we'll multiply both sides by 0.1.
13:39So on the left hand side we have
13:42233 equals xar minus 3 which is just the
13:46leftover of our
13:48numerator. And we come up with a sample
13:51mean xbar of 3.233.
13:57So therefore any sample of size
14:01225 with a sample mean greater than
14:073.233 would lead to a rejection of the
14:10null hypothesis assuming a constant
14:13sigma and the same alpha level. So if we
14:17went out again and we collected another
14:20sample, same 225 sample size, and let's
14:25say we got a mean of
14:313.19, what would happen
14:33then? Well, it's not greater than
14:373.233. Therefore, we would fail to
14:41reject the null hypothesis. See how this
14:44works? So what we've done is we've set
14:46up sort of a threshold mean value or
14:50sample mean value and anything above
14:53that assuming all this stays the same
14:56would lead to a rejection of the null
14:58hypothesis. Anything equal to or below
15:01that would make would mean we fail to
15:04reject the null hypothesis.

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