Question

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?

Answer 1

Answer:

a) The 90% confidence interval would be given (0.561;0.719).

b) $p_v =P(z>2.8)=1-P(z$

c) Using the significance level assumed $\alpha=0.01$ we see that $p_v$ 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

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

$p_o=0.5$ is the value that we want to test since the problem says majority    

$\alpha$ represent the significance level (no given, but is assumed)    

z would represent the statistic (variable of interest)    

$p_v$ 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

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

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

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.64 - 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.561$

$0.64 + 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.719$

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: $p \leq 0.5$  

Alternative Hypothesis: $p >0.5$  

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 $\hat p$ is significantly (different,higher or less) from a hypothesized value $p_o$".  

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  

$np_o =100*0.64=64>10$  

$n(1-p_o)=100*(1-0.64)=36>10$  

Calculate the statistic    

The statistic is calculated with the following formula:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o(1-p_o)}{n}}}$  

On this case the value of $p_o=0.5$ is the value that we are testing and n = 100.  

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=2.8$

The p value for the test would be:  

$p_v =P(z>2.8)=1-P(z$

Part c

Using the significance level assumed $\alpha=0.01$ we see that $p_v$ 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%.