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Answer:
The decision rule will be as follows: Reject the null hypothesis if test-statistic < -2.5 or fail to reject the
Now, Our test-statistic was -1.96, which is greater than -2.5. Thus, we will fail to reject the null hypothesis.
Step-by-step explanation:
We are given;
Population mean; μ = 432
Sample mean; x¯ = 426
Sample size; n = 24
Variance; Var = 225
Now, standard deviation from variance is; σ = √Var
Thus, σ = √225
σ = 15
Hypothesis is defined as;
Null hypothesis; H0: μ ≥ 432
Alternative hypothesis: Ha: μ < 432
Formula for the test statistic since sample size is less than 30 is;
t = (x¯ - μ)/(σ/√n)
Plugging in the relevant values;
t = (426 - 432)/(15/√24)
t = -1.96
Now,we are given significance level of 0.01; our DF = n - 1 = 24 - 1 = 23.
From t-table attached, we can see that at DF of 23 and significance value of 0.01, the critical value is 2.5
However, this is a left tailed test as the rejection region is to the left.
Thus, the critical value of a left tailed test is negative. Thus, in this case our critical value will be -2.5.
The decision rule will be as follows: Reject the null hypothesis if test-statistic < -2.5 or fail to reject the
Now, Our test-statistic was -1.96, which is greater than -2.5. Thus, we will fail to reject the null hypothesis.
