Question

A Pew Research Center poll of 2076 randomly selected adultsshowed that 91 % of them own cell phones. The following data results from a test ofthe claim that 92% of adults own cell phones.Test ofSample1p=0.92X1889vsN2076p≠0.92Sample p0.90992395% CI(0.897608,0.922238)Z-Value-1.69P-Value0.091$$ Use the normal distribution as an approximation to the binomial distribution. Use a 0.05 significance level and answer the following: What is the test statistic?

Answer 1

Answer: given below

Explanation:

we will start by carefully analyzing the question, as it looks tricky.

(a). Let us test the hypothesis, here we say that the proportion of adults own cell phones differs from 92 % at a 5 % significance level.

testing the null and alternative hypothesis gives;

Hₐ : p is not equal to 0.92

H₀ : p = 0.92

∴ this makes the hypothesis two-tailed

(b). From the MINI TAB output, the Z-test statistics is -1.69

(c). From the Stat Crunch output, the p- value for this test is 0.091

(d). The conclusion of this is that, the p - value is higher than 0.05 which is 0.091, that tells us that the null hypothesis doesn't get rejected at 5 % level of significance.

(e). In conclusion;

i will say that the result is not significant because there is no enough of evidence to tell that the proportion of adults own cell phones is different from 92 %.

cheers i hope this helps!!