After once again losing a football game to the college'ss arch rival, the alumni association conducted a survey to see if alumni were in favor of firing the coach. A simple random sample of 100 alumni from the population of all living alumni was taken. Sixty-four of the alumni in the sample were in favor of firing the coach. Let p represent the proportion of all living alumni who favor firing the coach

What is a 90% confidence interval for p? ( upper and lower limit to 4 decimal places)

Suppose the alumni association wished to see if the majority of alumni are in favor of firing the coach. To do this they test the hypotheses H0: p = 0.50 versus Ha: p > 0.50. What is the P-value for this hypothesis test? ( 4 decimal places)

Suppose the alumni association wished to conduct the test at a 1% significance level. What would their decision be? Based on that decision, what type of mistake could they have made?

c) Using the significance level assumed [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=64 represent were in favor of firing the coach

[tex]\hat p=\frac{64}{100}=0.64[/tex] estimated proportion for were in favor of firing the coach

[tex]p_o=0.5[/tex] is the value that we want to test since the problem says majority

[tex]\alpha[/tex] represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)

p= population proportion of Americans for were in favor of firing the coach

Part a

The confidence interval would be given by this formula

[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

For the 90% confidence interval the value of [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.

[tex]z_{\alpha/2}=1.64[/tex]

And replacing into the confidence interval formula we got:

[tex]0.64 - 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.561[/tex]

[tex]0.64 + 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.719[/tex]

And the 90% confidence interval would be given (0.561;0.719).

Part b

We need to conduct a hypothesis in order to test the claim that the proportion exceeds 50%(Majority). :

Null Hypothesis: [tex]p \leq 0.5[/tex]

Alternative Hypothesis: [tex]p >0.5[/tex]

We assume that the proportion follows a normal distribution.

This is a one tail upper test for the proportion of union membership.

The One-Sample Proportion Test is "used to assess whether a population proportion [tex]\hat p[/tex] is significantly (different,higher or less) from a hypothesized value [tex]p_o[/tex]".

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

[tex]np_o =100*0.64=64>10[/tex]

[tex]n(1-p_o)=100*(1-0.64)=36>10[/tex]

Calculate the statistic

The statistic is calculated with the following formula:

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o(1-p_o)}{n}}}[/tex]

On this case the value of [tex]p_o=0.5[/tex] is the value that we are testing and n = 100.

[tex]z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=2.8[/tex]

The p value for the test would be:

[tex]p_v =P(z>2.8)=1-P(z<2.8)=0.0026[/tex]

Part c

Using the significance level assumed [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.