Question

In an annual report to investors, an investment firm claims that the share price of one of their bond funds had very little variability. The report shows the average price as $⁢14.00 with a variance of 0.14. One of the investors wants to investigate this claim. He takes a random sample of the share prices for 29 days throughout the last year and finds that the standard deviation of the share price is 0.2266. Can the investor conclude that the variance of the share price of the bond fund is different than claimed at α=0.05? Assume the population is normally distributed.

Step 1 of 5: State the null and alternative hypotheses. Round to four decimal places when necessary.

Step 2 of 5: Determine the critical value(s) of the test statistic. If the test is two-tailed, separate the values with a comma. Round your answer to three decimal places.

Step 3 of 5: Determine the value of the test statistic. Round your answer to three decimal places.

Step 4 of 5: Make the decision

Step 5 of 5: What is the conclusion?

Answer 1

Solution:

Step 1

Null hypothesis: H0: The population variance of price of bond funds is 0.14.

Alternative hypothesis: Ha: The population variance of price of bound funds is different than 0.14.

H0: σ2 = 0.1400 versus Ha: σ2 ≠ 0.1400

This is a two tailed test.

Step 2

We are given

Sample size = n = 29

Degrees of freedom = n – 1 = 29 – 1 = 28

Level of significance = α = 0.05

So, critical values by using Chi square table are given as below:

Critical values: 15.308, 44.461

Step 3

The test statistic formula is given as below:

Chi square = (n – 1)*S^2/σ^2

We are given

σ^2 = 0.14

S = 0.2266

Chi square = (29 – 1)*0.2266^2 / 0.14

Chi square = 28*0.2266^2 / 0.14

Chi square = 10.26951

Test statistic = 10.270

Step 4

The test statistic value 10.270 is not lies between critical values 15.308 and 44.461, so we reject the null hypothesis.

Step 5

There is sufficient evidence to conclude that the population variance of price of bound funds is different than 0.14.