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$Question 4 1pts You wish to test the following claim (H_(a)) at a significance level of alpha=0.05. {:[H_(o):p=0.42],[H_(a):p > 0.42]:} You obtain a sample of size n=143 in which there are 73 successful observations. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution. What is the p-value? Round to 4 decimal places.$

Question $4$ $\newline$ $1 \mathrm{pts}$ $\newline$ You wish to test the following claim $\left(\mathrm{H}_{\mathrm{a}}\right)$ at a significance level of $\alpha=0.05$ . $\newline$ $\begin{array}{l} \mathrm{H}_{\mathrm{o}}: p=0.42 \\ \mathrm{H}_{\mathrm{a}}: p>0.42 \end{array}$ $\newline$ You obtain a sample of size $n=143$ in which there are $73$ successful observations. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution. $\newline$ What is the p-value? Round to $4$ decimal places.

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Q. Question

4

\newline

1 \mathrm{pts}

\newline

You wish to test the following claim

\left(\mathrm{H}_{\mathrm{a}}\right)

at a significance level of

\alpha=0.05

\newline

\begin{array}{l} \mathrm{H}_{\mathrm{o}}: p=0.42 \\ \mathrm{H}_{\mathrm{a}}: p>0.42 \end{array}

\newline

You obtain a sample of size

n=143

in which there are

73

successful observations. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution.

\newline

What is the p-value? Round to

4

decimal places.

Question Prompt: Question prompt: What is the $p$ -value for the hypothesis test with a sample size of $n=143$ and $73$ successful observations, without using continuity correction and using the normal distribution approximation?
Calculate Sample Proportion: Calculate the sample proportion $\hat{p}$ by dividing the number of successful observations by the sample size: $\hat{p} = \frac{73}{143}$ .
Find Standard Error: Find the standard error ( $SE$ ) using the formula $SE = \sqrt{(p*(1-p))/n}$ , where $p$ is the hypothesized population proportion from $H_0$ and $n$ is the sample size. $SE = \sqrt{(0.42*(1-0.42))/143}$ .
Calculate Z-Score: Calculate the z-score using the formula $z = \frac{\hat{p} - p}{SE}$ . Substitute the values: $z = \frac{\frac{73}{143} - 0.42}{SE}$ .
Perform Calculations: Perform the calculations: $\hat{p} = \frac{73}{143} \approx 0.5105$ , $SE = \sqrt{\left(0.42\times(1-0.42)\right)/143} \approx 0.0424$ , $z = \frac{(0.5105 - 0.42)}{0.0424} \approx 2.1321$ .
Find P-Value: Find the p-value by looking up the $z$ -score in the standard normal distribution table or using a calculator. The p-value corresponds to the area to the right of the $z$ -score since $H_a$ is $p > 0.42$ .
Find P-Value: Find the p-value by looking up the z-score in the standard normal distribution table or using a calculator. The p-value corresponds to the area to the right of the z-score since $H_a$ is $p > 0.42$ .The p-value is found to be approximately $0.0166$ after looking up a z-score of $2.1321$ in the standard normal distribution table.