Question

The total precipitation in Normal City in January is approximately normally distributed with a mean of 3.2 inches, and a standard deviation of 0.69. A researcher claims that the standard deviation, o, of the total March precipitation in recent years is less than 0.69. A random sample of 10 precipitation measurements from the last three decades is considered and the mean of these measurements is found to be 3.2 with a standard deviation of 0.39. Is there enough evidence to support the researcher's claim at the 0.1 level of significance? Perform a one-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. The null hypothesis р x S co The alternative hypothesis: . The type of best statistics (Choose one) D-OOSO 020 D> The value of the test statistie: (Round to at least three decimal places) 0 ? The p-value: (Round to at least three decimal places) 0

Answer 1

Solution:

Given:

Sample size = n = 10

Sample standard deviation = s = 0.39

Population standard deviation = $\sigma = 0.69$

Claim: The standard deviation $\sigma$ of the total March precipitation in recent years is less than 0.69.

Level of significance = 0.1

Part i) Null hypothesis:

$H_{0}: \sigma = 0.69$

Part ii) Alternative hypothesis:

$H_{1}: \sigma < 0.69$

Part iii) Type of test statistic:

We use Chi-square test of variance ( standard deviation)

Part iv) Test statistic value:

$\chi ^{2}=\frac{(n-1) \times s^{2}}{\sigma^{2}}$

$\chi ^{2}=\frac{(10-1) \times 0.39^{2}}{0.69^{2}}$

$\chi ^{2}=\frac{ 9 \times 0.1521 }{0.4761}$

$\chi ^{2}=2.875$

Part v) P-value:

Use following Excel command:

=CHISQ.DIST(X² , df,cumulative)

where

df = n - 1 = 10 - 1 = 9

thus

=CHISQ.DIST(2.875,9,TRUE)

=0.031

Thus P-value = 0.031

Part vi) Conclusion:
Decision Rule:
Reject null hypothesis H0, if P-value < 0.10 level of significance, otherwise we fail to reject H0

Since P-value = 0.031 < 0.10 level of significance, we reject null hypothesis H0.

Thus at 0.10  level of significance, we have sufficient evidence to support the claim that the standard deviation $\sigma$ of the total March precipitation in recent years is less than 0.69.