Question

A custodian wishes to compare two competing floor waxes to decide which one is best. He believes that the mean of WaxWin is less than the mean of WaxCo.
In a random sample of 35 floors of WaxWin and 53 of WaxCo. WaxWin had a mean lifetime of 25.7 and WaxCo had a mean lifetime of 29.9.
The population standard deviation for WaxWin is assumed to be 9.8 and the population standard deviation for WaxCo is assumed to be 10.8.
Perform a hypothesis test using a significance level of 0.05 to help him decide.
Let WaxWin be sample 1 and WaxCo be sample 2.
The correct hypotheses are:

• H0:μ1≤μ2H0:μ1≤μ2
  HA:μ1>μ2HA:μ1>μ2(claim)
• H0:μ1≥μ2H0:μ1≥μ2
  HA:μ1<μ2HA:μ1<μ2(claim)
• H0:μ1=μ2H0:μ1=μ2
  HA:μ1≠μ2HA:μ1≠μ2(claim)

Since the level of significance is 0.05 the critical value is -1.645
The test statistic is: (round to 3 places)
The p-value is: (round to 3 places)

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ1 >= μ2
Alternative Hypothesis, Ha: μ1 < μ2

Rejection Region
This is left tailed test, for α = 0.05
Critical value of z is -1.645.
Hence reject H0 if z < -1.645

Pooled Variance
sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(96.04/35 + 116.64/53)
sp = 2.2237

Test statistic,
z = (x1bar - x2bar)/sp
z = (25.7 - 29.9)/2.2237
z = -1.889

P-value Approach
P-value = 0.029
As P-value >= 0.05, fail to reject null hypothesis.