An orange grower is introducing a new variety of orange. The grower claims that the new oranges have a mean pH of 5 . To test the claim, an independent researcher chose a random sample of 22 of the new oranges, finding they had a sample mean pH of 5.1 with a sample standard deviation of 0.8 . Assume that the population of $\mathrm{pH}$ values of the new oranges is approximately normally distributed.
Complete the parts below to perform a hypothesis test to see if there is enough evidence, at the 0.10 level of significance, to reject that $\mu$, the mean pH of the new oranges, is 5 .
(a) State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$ that you would use for the test.
\[
\begin{array}{l}
H_{0}: \square \\
H_{1}: \square
\end{array}
\]
(b) Perform a hypothesis test. The test statistic has a $t$ distribution (so the test is a " $t$ test"). Here is some other information to help you with your test.
- $t_{0.05}$ is the value that cuts off an area of 0.05 in the right tail.
- The value of the test statistic is given by $t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}$.
Student's t Distribution
Step 1: Enter the number of degrees
of freedom.
Step 2: Select one-tailed or two-talied.
Answer
(a) The null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$ for the test are as follows: \[ \begin{array}{l} H_{0}: \mu = 5 \\ H_{1}: \mu \neq 5 \end{array} \] The null hypothesis states that the mean pH of the new oranges is 5, as claimed by the grower. The alternative hypothesis states that the mean pH is not equal to 5, indicating that the grower's claim is incorrect. (b) To perform the hypothesis test, we will use the given sample mean, sample standard deviation, and sample size to calculate the test statistic using the formula provided: \[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \] Given: - Sample mean ($\bar{x}$) = 5.1 - Population mean ($\mu$) = 5 (as per $H_{0}$) - Sample standard deviation (s) = 0.8 - Sample size (n) = 22 Plugging in the values, we get: \[ t = \frac{5.1 - 5}{\frac{0.8}{\sqrt{22}}} = \frac{0.1}{\frac{0.8}{\sqrt{22}}} = \frac{0.1}{0.1706} \approx 0.586 \] The degrees of freedom (df) for a t-test is equal to the sample size minus one, so df = 22 - 1 = 21. Since we are testing at the 0.10 level of significance and the alternative hypothesis is two-tailed, we need to find the critical t-value that cuts off an area of 0.05 in each tail (0.10 total). Using a t-distribution table or calculator, we find that $t_{0.05}$ with 21 degrees of freedom is approximately 1.721. Since the calculated t-value (0.586) is less than the critical t-value (1.721), we do not reject the null hypothesis. There is not enough evidence at the 0.10 level of significance to conclude that the mean pH of the new oranges is different from 5.
Step 1 ：(a) The null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$ for the test are as follows: \[ \begin{array}{l} H_{0}: \mu = 5 \\ H_{1}: \mu \neq 5 \end{array} \] The null hypothesis states that the mean pH of the new oranges is 5, as claimed by the grower. The alternative hypothesis states that the mean pH is not equal to 5, indicating that the grower's claim is incorrect. (b) To perform the hypothesis test, we will use the given sample mean, sample standard deviation, and sample size to calculate the test statistic using the formula provided: \[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \] Given: - Sample mean ($\bar{x}$) = 5.1 - Population mean ($\mu$) = 5 (as per $H_{0}$) - Sample standard deviation (s) = 0.8 - Sample size (n) = 22 Plugging in the values, we get: \[ t = \frac{5.1 - 5}{\frac{0.8}{\sqrt{22}}} = \frac{0.1}{\frac{0.8}{\sqrt{22}}} = \frac{0.1}{0.1706} \approx 0.586 \] The degrees of freedom (df) for a t-test is equal to the sample size minus one, so df = 22 - 1 = 21. Since we are testing at the 0.10 level of significance and the alternative hypothesis is two-tailed, we need to find the critical t-value that cuts off an area of 0.05 in each tail (0.10 total). Using a t-distribution table or calculator, we find that $t_{0.05}$ with 21 degrees of freedom is approximately 1.721. Since the calculated t-value (0.586) is less than the critical t-value (1.721), we do not reject the null hypothesis. There is not enough evidence at the 0.10 level of significance to conclude that the mean pH of the new oranges is different from 5.
