Questions: A nutritionist claims that the mean tuna consumption by a person is 3.5 pounds per year. A sample of 90 people shows that the mean tuna consumption by a person is 3.4 pounds per year. Assume the population standard deviation is 1.11 pounds. At α=0.07, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. A. H0: μ>3.5 B. H0: μ ≤ 3.5 Ha: μ ≤ 3.5 Ha: μ>3.5 C. H0: μ>3.4 Ha: μ ≤ 3.4 D. H0: μ=3.5 Ha: μ ≠ 3.5 E. H0: μ ≠ 3.4 Ha: μ=3.4 F. H0: μ ≤ 3.4 Ha: μ>3.4 (b) Identify the standardized test statistic, z. z= (Round to two decimal places as needed.) (c) Find the P-value. Hint: Keep in mind that P-value depends on the type of test. Left-tailed test: P-value=Area Right-tailed test: P-value=1-Area

A nutritionist claims that the mean tuna consumption by a person is 3.5 pounds per year. A sample of 90 people shows that the mean tuna consumption by a person is 3.4 pounds per year. Assume the population standard deviation is 1.11 pounds. At α=0.07, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. A. H0: μ>3.5 B. H0: μ ≤ 3.5 Ha: μ ≤ 3.5 Ha: μ>3.5 C. H0: μ>3.4 Ha: μ ≤ 3.4 D. H0: μ=3.5 Ha: μ ≠ 3.5 E. H0: μ ≠ 3.4 Ha: μ=3.4 F. H0: μ ≤ 3.4 Ha: μ>3.4 (b) Identify the standardized test statistic, z. z= (Round to two decimal places as needed.) (c) Find the P-value. Hint: Keep in mind that P-value depends on the type of test. Left-tailed test: P-value=Area Right-tailed test: P-value=1-Area
Transcript text: A nutritionist claims that the mean tuna consumption by a person is 3.5 pounds per year. A sample of 90 people shows that the mean tuna consumption by a person is 3.4 pounds per year. Assume the population standard deviation is 1.11 pounds. At $\alpha=0.07$, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. A. $\mathrm{H}_{0}: \mu>3.5$ B. $H_{0}: \mu \leq 3.5$ $H_{a}: \mu \leq 3.5$ $\mathrm{H}_{\mathrm{a}}: \mu>3.5$ C. $\mathrm{H}_{0}: \mu>3.4$ $\mathrm{H}_{a}: \mu \leq 3.4$ D. $\mathrm{H}_{0}: \mu=3.5$ $H_{a}: \mu \neq 3.5$ E. $\mathrm{H}_{0}: \mu \neq 3.4$ $\mathrm{H}_{\mathrm{a}}: \mu=3.4$ F. $\mathrm{H}_{0}: \mu \leq 3.4$ $\mathrm{H}_{\mathrm{a}}: \mu>3.4$ (b) Identify the standardized test statistic, z. $z=$ $\square$ (Round to two decimal places as needed.) (c) Find the P-value. Hint: Keep in mind that P-value depends on the type of test. Left-tailed test: P-value=Area Right-tailed test: $\mathbf{P}$-value=1-Area
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Solution

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Solution Steps

Step 1: Hypotheses

We set up the null and alternative hypotheses as follows:

  • Null Hypothesis: \( H_0: \mu = 3.5 \)
  • Alternative Hypothesis: \( H_a: \mu \neq 3.5 \)
Step 2: Standardized Test Statistic

To calculate the standardized test statistic \( z \), we use the formula: \[ z = \frac{X - \mu}{\sigma} \] where:

  • \( X = 3.4 \) (sample mean)
  • \( \mu = 3.5 \) (hypothesized population mean)
  • \( \sigma = \frac{1.11}{\sqrt{90}} \approx 0.1170 \) (standard error)

Calculating \( z \): \[ z = \frac{3.4 - 3.5}{0.1170} \approx -0.85 \]

Step 3: P-value Calculation

Since this is a two-tailed test, we calculate the P-value as: \[ P\text{-value} = 2 \times (1 - P(Z \leq z)) \] Using the cumulative distribution function (CDF) for the standard normal distribution, we find: \[ P\text{-value} \approx 1.6047 \]

Step 4: Decision

We compare the P-value to the significance level \( \alpha = 0.07 \):

  • Since \( P\text{-value} \approx 1.6047 > 0.07 \), we fail to reject the null hypothesis.

Final Answer

The null hypothesis cannot be rejected, indicating that there is not enough evidence to dispute the nutritionist's claim.

\(\boxed{H_0 \text{ cannot be rejected}}\)

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