Question

4 5 2 2 0.5 2.5 1 2 3.5 6 1.5 2 1 6.5 7 3...

4 5 2 2 0.5 2.5 1 2 3.5 6 1.5 2 1 6.5 7 3 3 3 1 1.54 6.5 3 2 4 1 2 2 2.5 4
7 1 5 1 2 3 5 7 1 4 6 1.5 2 2 2 2 2 0.5 2 5 4 2 1 2.5 6 2.5 1 4 4

Use the data set provided in the main page of the experience for this application.
1. Complete the full hypothesis testing procedure to determine if students at TCC watch less television than Americans in general. Use the fact that Americans watch 5.5 hours of television per day with a standard deviation of 1 hour per day. Include the following in your report:
a) Hypotheses using correct notation (2 points)
b) type of test (left, right, two-tailed) (1 point), distribution used (normal or t) (1 point), reasons why (2 points)
c) level of significance, choose one based on how serious the study is (2 points)
d) Sample statistics (sample size and sample mean or sample proportion) (2 points)
e) P-Value (2 point) (Round answer to four decimal places)
f) Interpretation of P-value (Use Definition of P-Value but be specific to this context) (1 point)
g) Drawing or graph of P-value. (2 points)
h) Decision (2 point)
i) Conclusion in complete sentences. (3 points)

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Answer #1

First of all we find the mean of the sample data using the formula :

ar{X}= rac{sum_{i=1}^{n}X_i}{n}

We get,

Y-2,983

Since, we have the population standard deviation known to us, Normal distribution will be used.

Now, we conduct the Z-test for one population mean as :

The provided sample mean is X 2.983 and the known population standard deviation is σ 1, and the sample size is n58. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested Ho: μ > 5.5 Ha: μ < 5.5 This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a 0.05, and the critical value for a left- tailed test is =-1.64. The rejection region for this left-tailed test is R = {z : z <-1.64) The z-statistic is computed as follows: X o2.983 -5.5 σ/Vn _ _ 1/V58-=-19.169(4) Decision about the null hypothesis Since it is observed that z-19.169 < Zc1.64, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p 0, and since p 00.05, 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 mean μ is less than 55, at the 0.05 significance level

Graphically z-Test Results: t-stats =-19.169, p-value <.0001 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 20 18 16 14 12 -10 8 620 24

Hence, we can conclude that there is enough evidence to say that Students at TCC watch less television than Americans in general.

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