He value represented by the 2 in 423 is times the value represented by the 2 in 72.
1 answer:
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Answer:
No, there is not enough evidence at the 0.02 level to support the strategist's claim.
Step-by-step explanation:
We are given that a political study took a sample of 1300 voters in the town and found that 57% of the residents favored construction.
And, a political strategist wants to test the claim that the percentage of residents who favor construction is more than 54%, i.e;
Null Hypothesis, : p = 0.54 {means that the percentage of residents who favor construction is 54%}
Alternate Hypothesis, : p > 0.54 {means that the percentage of residents who favor construction is more than 54%}
The test statistics we will use here is;
T.S. = ~ N(0,1)
where, p = actual percentage of residents who favor construction = 0.54
= percentage of residents who favor construction in a sample of
1300 voters = 0.57
n = sample of voters = 1300
So, Test statistics =
= 2.185
Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have insufficient evidence to reject null hypothesis.
Therefore, we conclude that the percentage of residents who favor construction is 54% and the strategist's claim is not supported.