Answer:
The lower bound of a 99% C.I for the proportion of defectives = 0.422
Step-by-step explanation:
From the given information:
The point estimate = sample proportion
= 0.55
At Confidence interval of 99%, the level of significance = 1 - 0.99
= 0.01
Then the margin of error
E = 0.128156
E ≅ 0.128
At 99% C.I for the population proportion p is:
= 0.55 - 0.128
= 0.422
Thus, the lower bound of a 99% C.I for the proportion of defectives = 0.422
Answer:
No, the sample evidence is not statistically significant (P-value = 0.125).
To reject the null hypothesis, the P-value has to be smaller than the significance level, so the significance level to reject the null hypothesis has to be 0.125 or higher.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the average weekly grocery bill per five-person family in San Antonio is significantly different from the national average.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample has a size n=50.
The sample mean is M=133.474.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.193.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size are:
This test is a two-tailed test, with 49 degrees of freedom and t=1.563, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.125) is bigger than the significance level (0.05), the effect is not significant.
The null hypothesis failed to be rejected.
There is not enough evidence to support the claim that the average weekly grocery bill per five-person family in San Antonio is significantly different from the national average.