Question

Suppose a university advertises that its average class size is 33 or less. A student organization is concerned that budget cuts have led to increased class sizes and would like to test this claim. A random sample of 39 classes was selected, and the average class size was found to be 36.6 students. Assume that the standard deviation for class size at the college is 9 students. Using α=0.05, determine the p-value for this test.

          Suppose a university advertises that its average class size is 33 or less. A student organization is concerned that budget cuts have led to increased class sizes and would like to test this claim. A random sample of 39 classes was selected, and the average class size was found to be 36.6 students. Assume that the standard deviation for class size at the college is 9 students. Using α=0.05, determine the p-value for this test.

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Added by Olga B.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Kari Hasz

10/10/2023

Step 1

6 Population mean (μ) = 33 Sample standard deviation (s) = 9 Sample size (n) = 39 t = (36.6 - 33) / (9 / √39) t = 3.6 / (9 / √39) t ≈ 3.6 / (9 / 6.2449979984) t ≈ 3.6 / 1.4422205106 t ≈ 2.496 Show more…

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Suppose a university advertises that its average class size is 33 or less. A student organization is concerned that budget cuts have led to increased class sizes and would like to test this claim. A random sample of 39 classes was selected, and the average class size was found to be 36.6 students. Assume that the standard deviation for class size at the college is 9 students. Using α=0.05, determine the p-value for this test.

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Assume that you are using a significance level of $\alpha=0.05$ to test the claim that $p>0.5$ and that your sample is a simple random sample of size $n=64$ a. Assuming that the true population proportion is $0.65,$ find the power of the test, which is the probability of rejecting the null hypothesis when it is false. (In the procedure below, we refer to $p=0.5$ as the "assumed" value, because it is assumed in the null hypothesis; we refer to $p=0.65$ as the "alternative" value, because it is the value of the population proportion used as an alternative to 0.5.) Use the following procedure and see the figure below. (FIGURE CAN'T COPY) Step 1: $\quad$ Using the significance level, find the critical $z$ value(s). (For a right-tailed test, there is a single critical $z$ value that is positive; for a left-tailed test, there is a single critical value of $z$ that is negative; and a two-tailed test will have a critical $z$ value that is negative along with another critical $z$ value that is positive.) Step $2: \quad$ In the expression for the test statistic below, substitute the assumed value of $p$ (used in the null hypothesis). Evaluate $1-p$ and substitute that result for the en- try of $q$. Also substitute the critical value(s) for $z$. Then solve for the sample statistic $\hat{p} .$ (If the test is two-tailed, substitute the critical value of $z$ that is positive, then solve for the sample statistic $\hat{p}$. Next, substitute the critical value of $z$ that is negative, and solve for the sample statistic $\hat{p} .$ A two-tailed test should therefore result in two different values of $\hat{p} .$ The resulting value(s) of $\hat{p}$ separate the region(s) where the null hypothesis is rejected from the region where we fail to reject the null hypothesis. $$z=\frac{\hat{p}-p}{\sqrt{\frac{p q}{n}}}$$ Step 3: $\quad$ The calculation of power requires a specific value of $p$ that is to be used as an alternative to the value assumed in the null hypothesis. Identify this alternative value of $p$ (not the value used in the null hypothesis), draw a normal curve with this alternative value at the center, and plot the value(s) of $\hat{p}$ found in Step 2 Step 4: $\quad$ Refer to the graph in Step $3,$ and find the area of the new critical region bounded by the value(s) of $\hat{p}$ found in Step 2 . (Caution: When evaluating $\sqrt{p q / n},$ be sure to use the alternative value of $p,$ not the value of $p$ used for the null hypothesis.) This is the probability of rejecting the null hypothesis, given that the alternative value of $p$ is the true value of the population proportion. Because this is the probability of rejecting the false null hypothesis, it is the power of the test. b. Find $\beta,$ which is the probability of failing to reject the false null hypothesis. The value of $\beta$ is easily determined by finding the complement of the power.

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Transcript

00:01 Suppose the university advertises that its average class is 33 or less.

00:09 A student organization is concerned that budget cuts have led to increased class sizes and would like to test this claim.

00:19 A random sample of 39 classes was selected and the average class size was found to be 36 .6.

00:29 So this will be our null hypothesis.

00:33 So the average was 36 .6.

00:36 There we go.

00:37 Assume the standard deviation for class size at the college is 9 students and use a significance of .05.

00:45 Determine the p -value for this test.

00:48 So to calculate the p -value, let's calculate a test statistic.

00:54 And we're going to use a z test because we have the population standard deviation.

01:00 We also have a sample size large enough to warrant a z test.

01:05 Our alternative hypothesis is that the mu is greater than 33.

01:10 And our z test statistic is going to be 36 .6 minus 33 divided by 9 over the square root of 39.

01:23 And that is 2 .497.

01:28 So we could round our test to...

01:43 So let's go ahead and round that to 2 .50.

01:48 Now to find the p -value, we're going to go to our...we can use our standard normal distribution table.

02:00 So there's a lot of information here about a t -test and a t -table, but you shouldn't use a t -test for this...

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