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Your answer is incorrect.
- (c): Your answer is incorrect.
- (d): Your answer is incorrect.
The Journal de Botanique reported that the mean height of Begonias grown while being treated with a particular nutrient is 36 centimeters. To check whether this is still accurate, heights are measured for a random sample of 21 Begonias grown while being treated with the nutrient. The sample mean and sample standard deviation of those height measurements are 35 centimeters and 9 centimeters, respectively.
Assume that the heights of treated Begonias are approximately normally distributed. Based on the sample, can it be concluded that the population mean height of treated begonias, $\mu$, is different from that reported in the journal? Use the 0.05 level of significance.
Perform a two-tailed test. Then complete the parts below.
Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.)
(a) State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$.
\[
\begin{array}{l}
H_{0}: \mu=36 \\
H_{1}: \mu \neq 36
\end{array}
\]
(b) Determine the type of test statistic to use.
(c) Find the value of the test statistic. (Round to three or more decimal places.)
\[
-0.482
\]
(d) Find the $p$-value. (Round to three or more decimal places.)
\[
0.634
\]
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Answer
The final answer is: The test statistic (z-score) is \(\boxed{-0.509}\) and the p-value is \(\boxed{0.611}\). Since the p-value is greater than the significance level of 0.05, we fail to reject the null hypothesis. Therefore, we do not have enough evidence to conclude that the population mean height of treated begonias is different from that reported in the journal.
Step 1 ：State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$. The null hypothesis is that the population mean is equal to the reported mean (36 cm), and the alternative hypothesis is that the population mean is not equal to the reported mean. So, we have \(H_{0}: \mu=36\) and \(H_{1}: \mu \neq 36\).
Step 2 ：Determine the type of test statistic to use. Since we are dealing with a sample size greater than 30 and we know the standard deviation, we should use a z-test.
Step 3 ：Calculate the test statistic. The formula for the z-score is (sample mean - population mean) / (standard deviation / sqrt(sample size)). Using the given values, the z-score is \(-0.509\).
Step 4 ：Find the p-value. The p-value is the probability that we would observe a result as extreme as the test statistic, assuming the null hypothesis is true. The calculated p-value is \(0.611\).
Step 5 ：Since the p-value is greater than the significance level of 0.05, we fail to reject the null hypothesis. This means that we do not have enough evidence to conclude that the population mean height of treated begonias is different from that reported in the journal.
Step 6 ：The final answer is: The test statistic (z-score) is \(\boxed{-0.509}\) and the p-value is \(\boxed{0.611}\). Since the p-value is greater than the significance level of 0.05, we fail to reject the null hypothesis. Therefore, we do not have enough evidence to conclude that the population mean height of treated begonias is different from that reported in the journal.
