Question

1. A recent national survey stated 70% of college students said they get less than the recommended amount of sleep every night. A statistician decides to test this claim against the suspicion that the percentage is too high. The statistician randomly sampled 1500 college students from the population of college students and determines that 1020 college students stated they don’t get the recommended amount of sleep. Perform a hypothesis test to answer the question: Do the sample results support the statistician’s claim at

α = 5 %?

1. State the hypotheses (in words & symbols):

H0:

Ha:

2. Determine the model to test H0

Can one use a normal distribution for the Sampling Distribution model to perform this test? Please explain. If not, then what distribution could you use?

b) Using the appropriate notation, state the MEAN & STD ERROR of the sampling distribution: Mean:

STD Error:

3. Determine the Decision Rule:

State the decision rule:

4. Analyze the sample data: Using the appropriate notation, state the sample result of the test, p-value and the test statistic.

sample result (use appropriate notation) is:

p-value ( to 4 decimal places ) is:

test statistic (use appropriate notation) is:   

Hypothesis Testing Model:

5. State the Conclusion:

Do you agree with the statistician suspicion at α = 5%? YOU MUST EXPLAIN YOUR ANSWER OTHERWISE: No credit will be given.

Answer 1

The following information is provided: The sample size is N=1500, the number of favorable cases is X = X=1020, and the sample proportion is pˉ​=X/N​=1020/1500​=0.68, and the significance level is α=0.05

1. State the hypotheses (in words & symbols):

H0: The percentage of students not getting enough sleep is 70%

Ha: The percentage of students not getting enough sleep is significantly less than 70%

This corresponds to a left-tailed test, for which a z-test for one population proportion needs to be used.

2. Can one use a normal distribution for the Sampling Distribution model to perform this test?

Yes, because sample size is large enough 1500 and they are college students, representative of college students population and hence, normal.

Mean: pbar = 0.68

STD Error: sqrt(0.68*0.32/1500) =

3. State the decision rule:

Based on the information provided, the significance level is α=0.05, and the critical value for a left-tailed test is z_c​=−1.64.

So, Decision rule is z_cal > 1.64 or p-value < 0.05

4. sample result (use appropriate notation) is: 0.012

The z-statistic is computed as follows:

​=−1.69

5. p-value ( to 4 decimal places ) is:

Using the P-value approach: The p-value is p=0.0455, and since p=0.0455<0.05, it is concluded that the null hypothesis is rejected.

6. Do you agree with the statistician suspicion at α = 5%?

z_cal = −1.69 < z_c​=−1.64, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value isp=0.0455, and since p=0.0455<0.05, it is concluded that the null hypothesis is rejected.

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population percentage of students having less sleep p_o is less than 70% ​, at the α=0.05 significance level.

Yes, we agree with the statistician.

Graphically

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