Answer:
The test statistic is z = [tex]\frac{x-u}{σ/√n}[/tex]
the z- value is 2.25 > 1.96 at 5% level of significance
we will not accepted null hypothesis
the mean (x⁻) and μ do not differ significantly
Step-by-step explanation:
Step 1:-
Given n = 250 and
population mean(μ ) = 3325 and
standard deviation of population (σ )= 525
mean of the sample ( x ) = 3400
i) Null hypothesis (H0) : (μ ) = 3325
ii) alternative hypothesis(H1) :(μ )≠ 3325
iii) The level of significance ∝ =0.05 = 1.96
Step 2 :-
The test statistic z = [tex]\frac{x-u}{σ/√n}[/tex]
now substitute all these values , we get
[tex]z = \frac{3400-3325}{525/√250}[/tex]
on simplification
z = 2.25
The tabulated value z = 1.96 at 5% level of significance (from z- table)
The calculated value z = 2.25 > tabulated value 1.96 at 0.05 level of significance
There fore the null hypothesis is rejected at 5% level of significance.
Therefore is not statistically significant
conclusion:-
The null hypothesis is rejected at 5% level of significance.
Therefore that is not statistically significant
