Solve the inequality. Check your solutions. 5t +6>21
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Answer:
Given:
For battery k(sample 1):
Sample size, n = 75
Sample mean, X' = 9.41
Standard deviation = 0.723
Mean u1 =???
For battery Q(sample 2):
Sample size, n = 49
Sample mean, X' = 9.69
Standard deviation = 0.257
Mean u2 =???
Significance level = 0.05
1) The null and alternative hypotheses, will be given as:
H0 : u1 - u2 = 0
H1 : u1 - u2 ≠ 0
2) This is a two tailed test.
For the test statistic, we have :
≈ 3.02
The Zo = - 3.02
3) For the p value, using excel, we have:
2 * P(Z > (-3.018))
= 2 * NORMDIST (-3.018, 0, 1, true)
= 0.00254
For critical value:
Zcritical = Za/2 = 0.05/2
Zcritical = 0.025
We have alternative hypothesis,
H1 : u1 - u2 ≠ 0.
Therefore we reject null hypothesis H0 if Zo > Zcritical
4) We reject null hypothesis H0 if Zo > Zcritical. At 0.05 level of significance, there is not enough evidence to conclude that mean u1 and mean u2 are different.