Question

5.2. Recall, again, that the time to first pruning of basil plants is known to be normally distributed with an average of µ = 35 days and a standard deviation of σ = 3 days. This term, n another l ab with a stable environment where no construction i s going on, students i n Kathleen’s class are trying a new fertilizer combination on their standard sweet basil plants. They are interested in whether this new fertilizer is improving (decreasing) the average number of days to first pruning for their basil plants. 36 randomly chosen students observed that the average time to first pruning for their plants was x¯ = 34 days with a standard deviation of s = 2.4 days, and that their sample data looks fairly bell shaped. What is the probability of observing a time of first pruning of 34 days or less if nothing has changed from the old number of days to first pruning that was the norm? Does it appear that the new fertilizer combination is decreasing the number of days to first pruning?

a. If you wish to test whether there is merit to the supposition that this new fertilizer is the “ "cat's meow” and is helping the plants to grow faster, what null and alternative hypothesis would be of interest here? HINT: This test will have a left sided alternative.

c. State the z-test statistic value (from MINITAB).

d. Write the p-value statement (by hand) and its value (from MINITAB)

e. Does it appear that the new fertilizer is decreasing the number of days to first pruning, on average? Why or why not? Test with a level of significance of 5%.

g. State the t-test statistic value (from MINITAB) and the degrees of freedom (find by hand).

h. Write the p-value statement (by hand) and its value (from MINITAB)

i. For the 5% significance level, do you have significant evidence that the new fertilizer is decreasing the number of days to first pruning, on average? Why or why not?

Answer 1

Solution:-

5.2)

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u > 35
Alternative hypothesis: u < 35

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 0.40

c)

z = (x - u) / SE

z = - 2.50

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a z statistic test statistic of - 2.50 .

d) Thus the P-value in this analysis is 0.006.

Interpret results. Since the P-value (0.006) is less than the significance level (0.05), we have to reject the null hypothesis.

e)

From the above test we have sufficient evidence in the favor of the claim that that the new fertilizer is decreasing the number of days to first pruning, on average.