Instruction: Answer each word problem below. Support your...

Instruction: Answer each word problem below. Support your answer with a solution. 1. Identify the appropriate rejection region for the given level of significance, then illustrate. Show your solution (10 points) The average income of a customer service representative is P 23,000. A survey was conducted among 25 customer service representatives in the city and the mean is 24,000 and the standard deviation is P 5,000. Is there a significant difference between the population mean and sample mean at 5% significance level? 2. Compute for the test-statistic value. Show your solution (10 points) A teacher wants to find out if his teaching strategy is effective. A sample of 30 students took a 50-point examination and the average is 37 with a standard deviation of 7. Test at 5% significance level that the average of the students is greater than 35 which is the standard passing score. 3. Formulate a Null Hypothesis, then analyze the situations below and draw the appropriate conclusion. Support your answer with a solution. (10 points) According to a guidance counselor, an average student spent less than 7 hours each day for school works. A sample of 100 students in a certain school reveals that the mean hours spent of the students is 6.5 hours with a standard deviation of 0.8. Determine whether the claim of the guidance counselor is correct against the population mean of 7 hours with 95% confidence level.

Transcript

00:01 All right, so this problem site gives us three different examples of setting up hypothesis tests and finding and evaluating those tests at different levels of significance.

00:11 And so the first example is discussing the salaries of customer service representatives, and our null hypothesis is that the average customer service representative makes 23 ,000.

00:23 But to test this, a sample of 25 was taken, with a sample of 24 ,000, and a sample standard deviation of 5 ,000.

00:33 And so we're trying to test at the 5 % significance level if there's a statistically significant difference in the sample mean and the population mean, which would lead us to reject our null hypothesis that the population mean is 23 ,000.

00:49 And so because a sample size is 25, we're going to not rely on a z distribution and instead is a t distribution.

00:57 Anytime your population is 30 or fewer, you can assume it's a large enough sample to have a perfectly normal distribution.

01:05 And so instead we're going to be concerned with a t distribution with degrees of freedom equal to n minus 1.

01:11 So in this case, we have 24 degrees of freedom.

01:15 So that is what this column right here is showing.

01:19 And so we're interested in this row with 24 degrees of freedom.

01:23 And then because we're trying to see if there's a statistically significant difference, it doesn't matter in the direction.

01:29 We're using a two -sided distribution, and our significance level is 5%.

01:37 And so you can see here this chart is only showing a one -sided distribution.

01:41 And so instead of looking at the 0 .05, we're instead going to look at half of that 0 .025, so that on either side, this sums up to a total of 0 .05.

01:53 And so if the t statistic is greater than what has a probability, of 0 .025 in either direction.

02:03 That means that totally broadly speaking combined we have enough evidence to reject at the 5 % significance level.

02:11 So here this means that our t statistic has to be greater than or equal to 2 .06390.

02:22 So let's open a new page.

02:25 So a rejection range occurs when t is greater than or equal to 2 .06 930.

02:33 And t is calculated as our sample mean minus our population mean or null hypothesis divided by our standard sample deviation over the square root of our population size.

02:50 And so now we're just going to be plugging everything in.

02:53 So we have 24 ,000 minus 23 ,000 divided by 5 ,000 over the square root of 25.

03:04 So this is going to simplify it to 1 ,000 divided by 5 ,000 over 5.

03:14 So this 5 multiplies up.

03:16 We have 5 ,000 over 5 ,000, which equals 1.

03:25 So right now our test statistic is equal to 1, which is less than the required test statistic.

03:36 And this should be absolute value, sorry.

03:38 But it's less than the required test statistic to reject the null hypothesis.

03:43 So in this case, we're not rejecting the null hypothesis.

03:46 And so now let's put our rejection range in terms of our variable of interest.

03:52 So to do this, we take t, which is equal to 2 .0693.

04:00 It's pretty greater than it equal to.

04:02 And this is going to be equal to x bar minus 23 ,000, divided by, 5 ,000 over the squared event.

04:13 So this is basically saying, what does a sample mean that we need with a standard deviation of 5 ,000? sorry, we can put this in right away.

04:23 That would lead us to reject the null hypothesis.

04:26 And this will give us a rejection range.

04:29 And again, this is the absolute value...

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