Question

1. Perform the following hypothesis test using the critical value (traditional) method. Be sure to state the null and alternative hypotheses, identify the critical value, calculate the test statistic, compare the test statistic to the critical value, and state the conclusion.   Use English if you cannot write the mathematical symbols.

The mean lasting time of 2 competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax. Wax#1 had a mean time of 3 months, while Wax#2's mean was 2.9 months. The population standard deviations are 0.33 and 0.36, respectively. Does the data indicate that Wax#1 lasts longer than Wax#2? Test at a 5% level of significance.

2. Perform the following hypothesis test using the critical value (traditional) method. Be sure to state the null and alternative hypotheses, identify the critical value, calculate the test statistic, compare the test statistic to the critical value, and state the conclusion.   Use English if you cannot write the mathematical symbols.

The television habits of 30 children were observed. The sample mean was found to be 48.2 hours per week, with a standard deviation of 12.4 hours per week. Test the claim that the standard deviation for all children is no more than 16 hours per week. Use 10% confidence.

3. Perform the following hypothesis test using the critical value (traditional) method. Be sure to state the null and alternative hypotheses, identify the critical value, calculate the test statistic, compare the test statistic to the critical value, and state the conclusion.   Use English if you cannot write the mathematical symbols.

An airline claims that, on average, 5% of its flights are delayed each day. On a given day of 500 flights, 6.2% were delayed. Test the hypothesis that the average proportion of delayed flights is greater than 5%. Use α = 0.01.

Answer 1

Solution:-

1)

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

Null hypothesis: u₁< u₂
Alternative hypothesis: u₁ > u₂

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

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

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

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 0.1092
DF = 38

t = [ (x₁ - x₂) - d ] / SE

t = 0.916

t_critical = + 1.686

Rejection region is t > 1.686

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is thesize of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

Interpret results. Since the t-value(0.916) is less than the critical value (1.686), hence we cannot reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that Wax 1 lasts longer than Wax 2.