Question

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 9 begonias grown while being treated with the nutrient. the sample mean and sample standard deviation of those height measurements are 40 centimeters and 7 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, μ, 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 h₀ and the alternative hypothesis h₁. h₀:□ h₁:□ (b) determine the type of test statistic to use. select (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 no

Explanation:

Step1: State hypotheses

The null hypothesis $H_0$ is that the population mean $\mu$ is equal to the reported mean. The alternative hypothesis $H_1$ is that the population mean $\mu$ is not equal to the reported mean. So, $H_0:\mu = 36$ and $H_1:\mu
eq36$.

Step2: Determine test - statistic type

Since the population standard deviation $\sigma$ is unknown and we have a sample standard deviation $s$, and the population is approximately normally distributed, we use a t - test.

Step3: Calculate the t - test statistic

The formula for the t - test statistic is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$. Given $\bar{x} = 40$, $\mu = 36$, $s = 7$, and $n = 9$. Then $t=\frac{40 - 36}{7/\sqrt{9}}=\frac{4}{7/3}=\frac{12}{7}\approx1.714$.

Step4: Calculate the p - value

The degrees of freedom is $df=n - 1=9 - 1 = 8$. For a two - tailed t - test with $t = 1.714$ and $df = 8$, using a t - distribution table or software, the p - value is $2\times(1 - P(T<1.714))$ where $T$ has a t - distribution with $df = 8$. The p - value is approximately $0.125$.

Step5: Make a decision

Since the p - value ($0.125$) is greater than the significance level $\alpha=0.05$, we fail to reject the null hypothesis.

Answer:

(a) $H_0:\mu = 36$, $H_1:\mu
eq36$
(b) t - test
(c) $1.714$
(d) $0.125$
(e) No