Statistics(hypotheis testing )
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A researcher tested whether the mean number of days basic automobiles sit on dealers' lots is greater than 29 days. A sample of 30 dealers had a mean of 30.1 days. With a significance level of 0.05 and a population standard deviation of 3.8 days, a one-tailed t-test was conducted. The t-statistic was below the critical value so the null hypothesis that the mean is 29 days was not rejected, meaning there is not enough evidence to say the mean is greater than 29 days.
Statistics(hypotheis testing )
- 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
- 5. α = 0.05
- 8. α = 0.05 (level of significance) and the test is a one- tailed test Critical value Z = +1.65
- 9. Z = 30.1 – 29 3.8/ √ 30 = 1.59
- 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.