Questions: QUESTION 20 · 1 POINT Suppose a chef claims that her meatball weight is less than 4 ounces, on average. Several of her customers do not believe her, so the chef decides to do a hypothesis test, at a 10% significance level, to persuade them. She cooks 14 meatballs. The mean weight of the sample meatballs is 3.7 ounces. The chef knows from experience that the standard deviation for her meatball weight is 0.5 ounces. - H0: μ ≥ 4; Ha: μ<4 - α=0.1 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places? Provide your answer below: Test statistic =

QUESTION 20 · 1 POINT Suppose a chef claims that her meatball weight is less than 4 ounces, on average. Several of her customers do not believe her, so the chef decides to do a hypothesis test, at a 10% significance level, to persuade them. She cooks 14 meatballs. The mean weight of the sample meatballs is 3.7 ounces. The chef knows from experience that the standard deviation for her meatball weight is 0.5 ounces. - H0: μ ≥ 4; Ha: μ<4 - α=0.1 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places? Provide your answer below: Test statistic =
Transcript text: QUESTION 20 $\cdot$ 1 POINT Suppose a chef claims that her meatball weight is less than 4 ounces, on average. Several of her customers do not believe her, so the chef decides to do a hypothesis test, at a $10 \%$ significance level, to persuade them. She cooks 14 meatballs. The mean weight of the sample meatballs is 3.7 ounces. The chef knows from experience that the standard deviation for her meatball weight is 0.5 ounces. - $H_{0}: \mu \geq 4 ; H_{a}: \mu<4$ - $\alpha=0.1$ (significance level) What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places? Provide your answer below: Test statistic $=$ $\square$
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Solution

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Solution Steps

Step 1: Calculate the Standard Error

To calculate the standard error \( SE \) of the sample mean, we use the formula:

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{0.5}{\sqrt{14}} \approx 0.134 \]

Step 2: Calculate the Test Statistic

Next, we calculate the test statistic \( Z \) using the formula:

\[ Z = \frac{\bar{x} - \mu_0}{SE} = \frac{3.7 - 4}{0.134} \approx -2.24 \]

Step 3: Determine the P-value

For a left-tailed test, we find the probability associated with the calculated z-score:

\[ P = T(z) \approx 0.01 \]

Final Answer

The test statistic is

\[ \boxed{-2.24} \]

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