confidence interval level at 99% (step 2.) Below is my question and I will provide my answer for step one. Can someone please explain what I have to do next to complete the problem? PLEASE.
When analyzing the last digits of telephone numbers in Sandtown, it was found that among 1500 randomly selected phone numbers, 225 had zero as the last digit. If the digits are selected randomly, the proportion of zeroes should be 0.1 because from probability, 0 is one of the 10 possible digits.
-?Use the p-value method with a 0.01 level of significance to test the claim that the proportion of zeroes is not equal to 0.1.
-?Use the sample data to construct a 99% confidence interval to estimate the proportion of zeroes. What does the???confidence interval suggest about the claim that the proportion of zeroes is 0.1?
-Compare the results from the hypothesis test and confidence interval. Do they lead to the same conclusion? Why or why not?
My answer for part 1 :
to perform the statistical test , we first need to calculate the z stat as
The test statistic is a z-score (z) defined by the following equation.
z = (p - P) / ? , where p is sample proportion and P is population proportion
so p = 225/1500 = 0.15 and P = 0.1 and N = 1500 (given)
Compute the standard deviation (?) of the sampling distribution.
? = sqrt[ P * ( 1 - P ) / n ]
so sqrt(0.1*0.9/1500) = 0.0077
now putting this in the first equation we get
Z = (0.15-0.1)/0.0077 = 6.49
now we check the p value in the z table as
P ( Z<6.49 )=1
hence as we see that the p value is greater than 0.01, hence we fail to reject the null hypothesis and conclude that the proportion of zeros is not equal to 0.1
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