Mathematics, 05.12.2019 18:31 jksimmons2364

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?

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Answer from: marcy2215

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

b) p_v =P(z2.8)=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=6410

n(1-p_o)=100*(1-0.64)=3610

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(z2.8)=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%.

Answer from: Quest

1. 8 or -82. 14 or -143. -11, -3, 0, 7, 10

Answer from: Quest

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step-by-step explanation:

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