Question

3. The average return from equity mutual funds over a period was 2 %. It is desired to test if the returns from growth fund over the same period exceed that of the equity fund. A sample of 9 growth funds showed an average retun of 2.183 % . It is known from past data that the standard deviation of the age of returns for growth funds is 0.2%. If we test H0 : μ-2 against Ho : μ > 2 the data would lead us to

Please help! Only Multiple Choice.

Answer 1

Solution:-

3) (d) Rejection of null hypothesis at 10% level of significance but acceptance at 0.1% level of significance.

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

Null hypothesis: u = 2.0
Alternative hypothesis: u > 2.0

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.0667
DF = n - 1

D.F = 8
t = (x - u) / SE

t = 2.745

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 t statistic test statistic of 2.745

Thus the P-value in this analysis is 0.013

Interpret results.

Interpret results. Since the P-value (0.013) is less than the significance level (0.10), we have to reject the null hypothesis.Since the P-value (0.013) is greater than the significance level (0.001), we failed to reject the null hypothesis.

(d) Rejection of null hypothesis at 10% level of significance but acceptance at 0.1% level of significance.

4) (d) Rejection of null hypothesis at 10% level of significance but acceptance at 0.1% level of significance.

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

Null hypothesis: u = 35
Alternative hypothesis: u $\neq$ 35

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.5916

DF = n - 1

D.F = 8
t = (x - u) / SE

t = - 2.634

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.

Since we have a two-tailed test, the P-value is the probability that the t statistic having 8 degrees of freedom is less than -2.634 or greater than 2.634.

Thus, the P-value = 0.03

Interpret results. Interpret results. Since the P-value (0.03) is less than the significance level (0.10), we have to reject the null hypothesis.

Since the P-value (0.03) is greater than the significance level (0.001), we failed to reject the null hypothesis.

(d) Rejection of null hypothesis at 10% level of significance but acceptance at 0.1% level of significance.