In 1975, the standard deviation of the heights of residents of Townopolis was 5.
ID: 3262708 • Letter: I
Question
In 1975, the standard deviation of the heights of residents of Townopolis was 5.4 inches. A recent sample of 20 citizens of Townopolis had measured heights with a standard deviation of 6.6 inches. At =0.05=0.05, is there enough evidence to conclude that the standard deviation of heights is now not equal to 5.4 inches? What is the critical value? If there are two critical values, give the larger of the two. Round your answer to 2 decimal places.
Consider the following situation. In 1974, the standard deviation of the weights of residents of Townopolis was 5.6 pounds. A recent sample of 32 citizens of Townopolis had measured weights with a standard deviation of 4.8 pounds. At =0.05=0.05, is there enough evidence to conclude that the standard deviation of weights is now not equal to 5.6 pounds? What is the test statistic? Round your answer to 2 decimal places.
Consider the following situation. In 1987, the standard deviation of the ages of residents of Townopolis was 4.8 years. A recent sample of 30 citizens of Townopolis had ages with a standard deviation of 4.1 years. At =0.1=0.1, is there enough evidence to conclude that the standard deviation of ages is now not equal to 4.8 years? The two-tailed critical values are -17.71 and 17.71. The test statistic is 21.16. Can we reject the null hypothesis?
Consider the following situation. In 1979, the standard deviation of the ages of residents of Townopolis was 5.8 years. A recent sample of 29 citizens of Townopolis had ages with a standard deviation of 4.6 years. At =0.01=0.01, is there enough evidence to conclude that the standard deviation of ages is now not equal to 5.8 years? The two-tailed critical values are -12.46 and 12.46. The test statistic is 17.61. Can we conclude that the standard deviation of ages is now not equal to 5.8 years?
Explanation / Answer
Consider the following situation. In 1987, the standard deviation of the ages of residents of Townopolis was 4.8 years. A recent sample of 30 citizens of Townopolis had ages with a standard deviation of 4.1 years. At =0.1=0.1, is there enough evidence to conclude that the standard deviation of ages is now not equal to 4.8 years? The two-tailed critical values are -17.71 and 17.71. The test statistic is 21.16. Can we reject the null hypothesis?
The two-tailed critical values are -17.71 and 17.71. The test statistic is 21.16.
test statistic>two tailed critical values
21.16>17.71
Reject null hypothesis
there is enough evidence to conclude that the standard deviation of ages is now not equal to 4.8 years.
Consider the following situation. In 1979, the standard deviation of the ages of residents of Townopolis was 5.8 years. A recent sample of 29 citizens of Townopolis had ages with a standard deviation of 4.6 years. At =0.01=0.01, is there enough evidence to conclude that the standard deviation of ages is now not equal to 5.8 years? The two-tailed critical values are -12.46 and 12.46. The test statistic is 17.61. Can we conclude that the standard deviation of ages is now not equal to 5.8 years?
The two-tailed critical values are -12.46 and 12.46.
The test statistic is 17.61
17.61>12.46
Test statistic>critical value.
Reject Null Hypothesis
. we conclude that the standard deviation of ages is now not equal to 5.8 years