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In a survey,
43% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too high, so she randomly selected 250 pet owners and discovered that 104 of them spoke to their pet on the telephone. Does the veterinarian have a right to be skeptical? Use the
alpha=0.1 level of significance.
Because
np_(0)(1-p_(0))=◻◻10, the sample size is
◻
5% of the population size, and the sample

◻ the requirements for testing the hypothesis
◻ satisfied. (Round to one decimal place as needed.)

In a survey, $43 \%$ of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too high, so she randomly selected $250$ pet owners and discovered that $104$ of them spoke to their pet on the telephone. Does the veterinarian have a right to be skeptical? Use the $\alpha=0.1$ level of significance. $\newline$ Because $\mathrm{np}_{0}\left(1-\mathrm{p}_{0}\right)=\square \square 10$ , the sample size is $\square$ $5 \%$ of the population size, and the sample $\newline$ $\square$ the requirements for testing the hypothesis $\square$ satisfied. (Round to one decimal place as needed.)

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Interpret confidence intervals for population means

Full solution

Q. In a survey,

43 \%

of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too high, so she randomly selected

250

pet owners and discovered that

104

of them spoke to their pet on the telephone. Does the veterinarian have a right to be skeptical? Use the

\alpha=0.1

level of significance.

\newline

Because

\mathrm{np}_{0}\left(1-\mathrm{p}_{0}\right)=\square \square 10

, the sample size is

\square

5 \%

of the population size, and the sample

\newline

\square

the requirements for testing the hypothesis

\square

satisfied. (Round to one decimal place as needed.)

Calculate Proportion: Calculate the proportion of pet owners in the veterinarian's sample who talk to their pets on the telephone. $\newline$ Calculation: $\frac{104 \text{ pet owners}}{250 \text{ total pet owners}} = 0.416$
Set Hypotheses: Set up the null hypothesis ( $H_0$ ) and the alternative hypothesis ( $H_1$ ). $\newline$ $H_0$ : $p = 0.43$ (The proportion of all pet owners who talk to their pets on the telephone is $43\%$ ) $\newline$ $H_1$ : $p \neq 0.43$ (The proportion of all pet owners who talk to their pets on the telephone is not $43\%$ )
Calculate SE: Calculate the standard error (SE) using the formula $SE = \sqrt{p(1-p)/n}$ , where $p$ is the proportion under the null hypothesis and $n$ is the sample size. $\newline$ Calculation: $SE = \sqrt{0.43 \times (1 - 0.43) / 250} = \sqrt{0.43 \times 0.57 / 250} = \sqrt{0.2451 / 250} = \sqrt{0.0009804} = 0.0313$
Calculate z-score: Calculate the z-score for the sample proportion to see how far it deviates from the null hypothesis. $\newline$ Calculation: z = \frac{p_{\text{sample}} - p_{\text{null}}}{\text{SE}} = \frac{\(0\).\(416\) - \(0\).\(43\)}{\(0\).\(0313\)} = \(-0. $447$ \approx $-0$ . $45$
Determine Critical Value: Determine the critical z-value for a two-tailed test at $\alpha = 0.1$ . Since it's a two-tailed test, the critical z-values are approximately $\pm1.645$ .
Compare z-scores: Compare the calculated $z$ -score with the critical $z$ -values. Since $-0.45$ is not less than $-1.645$ or greater than $1.645$ , we do not reject the null hypothesis.

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