The Journal de Botanique reported that the mean height of Begonias grown while being treated with a particular nutrient is 45 centimeters. To check whether this is still accurate, heights are measured for a random sample of 12 Begonias grown while being treated with the nutrient. The sample mean and sample standard deviation of those height measurements are 52 centimeters and 10 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 0.10 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 $H_{1}$. \[ \begin{array}{l} H_{0}: \\ H_{1}: \end{array} \] (b) Determine the type of test statistic to use. (Choose one) (c) Find the value of the test statistic. (Round to three or more decimal places.) (d) Find the $p$-value. (Round to three or more decimal places.) (e) Can it be concluded that the mean height of treated Begonias is different from that reported in the journal? Yes ONO
Solution
Step 1 ：State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$. The null hypothesis is that the population mean height of treated begonias is 45 centimeters, as reported in the journal. The alternative hypothesis is that the population mean height of treated begonias is not 45 centimeters. So, we have: \[\begin{array}{l} H_{0}: \mu = 45 \\ H_{1}: \mu \neq 45 \end{array}\]
Step 2 ：Determine the type of test statistic to use. Since we know the sample standard deviation and not the population standard deviation, we will use a t-test.
Step 3 ：Calculate the test statistic. The formula for the t-test statistic is (sample mean - population mean) / (sample standard deviation / sqrt(sample size)). Using the given values, the test statistic is \(\boxed{2.425}\).
Step 4 ：Find the p-value. The p-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. The calculated p-value is \(\boxed{0.034}\).
Step 5 ：Since the p-value is less than the significance level of 0.10, we reject the null hypothesis. Therefore, it can be concluded that the mean height of treated Begonias is different from that reported in the journal. So, the answer is \(\boxed{\text{Yes}}\).
