Question

Does it take a different amount of time for seeds to germinate if they are near rock music that is continuously playing compared to being near classical music? The 43 seeds that were exposed to rock music took an average of 20 days to germinate. The standard deviation was 7 days. The 43 seeds that were exposed to classical music took an average of 25 days to germinate. The standard deviation for these seeds was 11 days. What can be concluded at the α=0.05 level of significance?

a. For this study, we should us
b. The null and alternative hypotneses would be
c. The test statistic = (please show your answer to 3 decimal places).
d. The p-value = (Please show your answer to 4 decimal places.)
e. The p-value is α f. Based on this, we should the null hypothesis.

Answer 1

To determine if there is a difference in the time it takes for seeds to germinate near rock music compared to classical music, we can perform an independent samples t-test. The hypotheses and test statistic can be calculated using the given data. The p-value can then be compared to the chosen significance level to draw a conclusion.

Step-by-step explanation:

To determine if there is a difference in the time it takes for seeds to germinate near rock music compared to classical music, we can perform a hypothesis test.

a. For this study, we should use an independent samples t-test because we are comparing the means of two independent groups (seeds exposed to rock music vs seeds exposed to classical music).

b. The null hypothesis (H0) would be that there is no difference in the average time to germinate between seeds exposed to rock music and seeds exposed to classical music. The alternative hypothesis (H1) would be that there is a difference in the average time to germinate between the two groups.

c. The test statistic can be calculated using the formula:

t = (mean1 - mean2) / √((sd1^2/n1) + (sd2^2/n2)) = (20 - 25) /√((7^2/43) + (11^2/43))

d. The p-value can be determined by comparing the test statistic to the appropriate t-distribution with degrees of freedom (df) equal to (n1 - 1) + (n2 - 1) = 43 - 1 + 43 - 1 = 84. We can use a t-table or statistical software to find the p-value associated with the calculated test statistic.

e. The p-value is the probability of obtaining a test statistic as extreme as the one calculated (or more extreme) if the null hypothesis is true. If the p-value is less than the chosen significance level (α = 0.05), we would reject the null hypothesis in favor of the alternative hypothesis.

f. Based on the p-value obtained, we should either reject or fail to reject the null hypothesis, depending on whether the p-value is less than or greater than α, respectively.