2. A researcher wishes to see if the mean number of
days that a basic, low-price, small automobile sits
on a dealer’s lot is 29. A sample of 30 automobile
dealers has a mean of 30.1 days for basic, low-
price, small automobiles. At α = 0.05, test the claim
that the mean time is greater than 29 days. The
standard deviation of the population is 3.8 days.
3. X = 30.1 ( Sample Mean )
n = 30 ( Sample Size)
μ = 29 ( Hypothesized Population Mean )
σ = 3.8 ( Population Standard Deviation )
α = 0.05 ( Level of Significance )
4. A researcher wishes to see if the mean number of
days that a basic, low-price, small automobile sits
on a dealer’s lot is 29. A sample of 30 automobile
dealers has a mean of 30.1 days for basic, low-
price, small automobiles. At α = 0.05, test the claim
that the mean time is greater than 29 days. The
standard deviation of the population is 3.8 days.
H0: μ = 29 and H1: μ > 29
10. 1.59 <1.65, and is not in the critical
region, the decision is not to reject
the null hypothesis.
z > zα
11. There is not enough evidence to support the claim
that the mean time is greater than 29 days.
12. the average weight of 100 randomly selected sacks
of rice is 48.54 kilos with a standard deviation of 20
kilos. Test the hypothesis at a 0.01 level of
significance that the true mean weight is less than
50 kilos.
Given:
X = 48.54 ( Sample Mean )
n = 100 ( Sample Size )
μ = 50 ( Hypothesized Population Mean )
s = 20 ( Standard Deviation )
α = 0.01 ( Level of Significance )
13. Step 1: H0: μ = 50 and H1: μ < 50
Step 2: significance level α = 0.01
Step 3: test statistic:
Step 4: critical regions: z < -zα which is z < -z0.01
Thus, we reject H0 if z < -2.33, otherwise we fail to
reject H0
Step 5: Z = 48.54 – 50
20/ √100
14. Step 6: statistical decision:
z < -zα
- 0.73 > -2.33
Since z = - 0.73 is not in the critical region, the null
hypothesis is not rejected.
Step 7: conclusion:
The test result does not provide sufficient evidence
to indicate that the true mean weight sack of rice is
less than 50 kilos.