Confidence intervals for the means provide an estimate for where the true mean lies.
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Answer:
We conclude that 50.
Step-by-step explanation:
We are given that x bar = 60, s = 12, n = 30, and alpha = 0.10
Also, Null Hypothesis, :
= 50
Alternate Hypothesis, :
50
The test statistics we will use here is;
T.S. = ~
where, X bar = sample mean = 60
s = sample standard deviation = 12
n = sample size = 30
So, Test statistics = ~
= 4.564
At 10% significance level, t table gives critical value of 1.311 at 29 degree of freedom. Since our test statistics is more than the critical value as 4.564 > 1.311 so we have sufficient evidence to reject null hypothesis as our test statistics will fall in the rejection region.
P-value is given by, P( > 4.564) = less than 0.05% {using t table}
Here also, P-value is less than the significance level as 0.05% < 10% , so we will reject null hypothesis.
Therefore, we conclude that 50.