2. Last year, the standard deviation of the ages of the students at UA was 1.8 y

Question

2. Last year, the standard deviation of the ages of the students at UA was 1.8 years. Recently, a sample of 31 students had a standard deviation of 2.1 years. We are interested in testing to see if there has been a significant change in the standard deviation of the ages of the students at UA. The null hypothesis is a. H>1.8 b. H18 C. ?>1.8 3. Refer to the information in Q2. The test statistic (chi-square value) is a. 22.04 b. 35.00 ?. 40.83 d. 6.20 4. Refer to the information in Q2. At 95% confidence the null hypothesis a. should be rejected b. should not be rejected c. should be revised d. None of these alternatives is correct. 5. The contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces. We are interested in testing to determine whether the variance of the population is significantly more than 0.003. The alternative hypothesis is a. S0003 b. S 0.003 ?2>0.003 ? 2-0.003 ?. 6. Refer to the information in Q5. The test statistic (chi-square value) is a. 1.2 b. 31.2 C. 30 d. 500 7, Refer to the information in Q5. At 95% confidence, the null hypothesis a. should be rejected b. should not be rejected c. should be revised d None of these alternatives is correct.

Explanation / Answer

#2. Option D
Null hypothesis: H0: sigma = 2.8

#3.
chi-square = (n-1)(s/sigma)^2
chi-square = (31 - 1)*(2.1/1.8)^2
chi-square = 40.83

Option C

#4.
The critical values of chi-square are 16.79 and 46.98
Hence null hypothesis should not be rejected
Option B

#5.
Option C

#6.
chi-square = (n-1)(s/sigma)^2
chi-square = (26 - 1)*(0.06^2/0.003)
chi-square = 30

Option C

#7.
The critical value of chi-square is 37.65
Null hypothesis should not be rejected
Option B

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