Use the survey data you collected last week to answer the following discussion q

Question

Use the survey data you collected last week to answer the following discussion question. Make a basic hypothesis about your data.

I hypothesize that the average survey respondent spends at least two hours watching reality TV a day.

Use the statistical techniques learned this week to test your hypothesis and explain your results. Were your results surprising? Why or why not?

1) State the Null & Alternative Hypothesis.

(2) Choose the appropriate statistical test.

(3) Find the critical value and/or p-value.

(4) Calculate the value of the test statistic.

(5) Compare the critical value with the test statistic and/or the p-value with the significance level (alpha-level).

(6) State the statistical decision and simple conclusion.

Please remember your ultimate goal in statistical inference: you are making a decision about a population based on a sample.

Survey data:

How many reality television shows do you watch per week?

Female responses- 28 (84.84%)            Male responses-5 (15.15%)

Answer Choices –

Responses –

0-1

54.55%

18

1-3

21.21%

7

3-5

12.12%

4

–

5-6

6.06%

2

–

6 or more

6.06%

2

TOTAL

33

Answer Choices –

Responses –

0-1

54.55%

18

1-3

21.21%

7

3-5

12.12%

4

–

5-6

6.06%

2

–

6 or more

6.06%

2

TOTAL

33

Explanation / Answer

1. State the hypotheses. The null hypothesis is the hypothesis of no difference. It states that on an average 50% of respondents spend atleast 2 hours of watching reality TV a day. The alternative hypothesis states that more than 50% of respondents spend atleast 2 hours of watching reality TV a day.

H0:p=0.5 Ha:p>0.5

2. State assumptions: Model-Random sampling, level of measurement is nominal, sampling distribution is normal. Based on assumptions, sampling distribution is Z distribution. Use 1-proportion Z test.

3. For right tailed test the critical Z value is +1.645.

4. The test statistic, Z=(phat-p)/sqrt[p(1-p)/n], where, phat is sample proportion, p is population proportion, n is sample size.

=(15/33-0.5)/sqrt[0.5(1-0.5)/33]

=-0.52

5. The test statistic does not fall in critical region (observed Z is not greater than critical Z). Therefore, fail to reject null hypothesis.

6. Conclude that there is insufficient sample evidence to suggest that more than 50% of respondents spend atleast 2 hours of watching reality TV a day.

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