MTH 245 One-Sample Test for the Mean
Order ID 53563633773 Type Essay Writer Level Masters Style APA Sources/References 4 Perfect Number of Pages to Order 5-10 Pages Description/Paper Instructions
MTH 245 One-Sample Test for the Mean
The methods in this section presume that the data fit at least one of the following two criteria:
- ππ β₯ 30, or 2. The original random variable is normally distributed.
If ππ < 30 and the original random variable is not normally distributed, we need to use bootstrapping or some other non-parametric method. (Bootstrapping will be covered in Section 8-3.) When the Central Limit Theorem Applies and ππ is Known In the rare case where we know the value of ππ (or can make an assumption of its value with reasonable certainty), we can make use of the Central Limit Theorem and calculate a p-value using the normal distribution. When the Central Limit Theorem Applies and ππ is Unknown Most of the time, we won’t know the value of ππ. In this case, we need to substitute the sample standard deviation π π for ππ. Unfortunately, π π is a biased estimator of ππ, so if we use it with the normal distribution, we’ll wind up with a p-value that is too large. We compensate for this by using the π‘π‘ distribution (aka Student’s t) instead of the normal distribution. The π‘π‘ distribution has greater variability than the standard normal distribution, and therefore compensates for the tendency for π π to underestimate ππ.
Comparison of the standard normal and π‘π‘ distributions (3 degrees of freedom)
The following table summarizes the foregoing discussion:
When⦠and⦠use⦠and the StatCrunch
menuβ¦
π₯π₯ is normal OR ππ β₯ 30
ππ is unknown t Distribution Stat Γ T-Stats
ππ is known Normal
Distribution Stat Γ Z-Stats
π₯π₯ is not normal AND ππ < 30
Bootstrapping/non-parametric methods
N/A
One-Sample Test for the Mean Using StatCrunch To use StatCrunch to conduct a one-sample hypothesis test for ππ when sample statistics (οΏ½Μ οΏ½π₯, π π , and ππ) are already available:
- Open a blank data table. 2. For the π‘π‘ distribution, click Stat Γ T Stats Γ One Sample Γ With
Summary. For the normal distribution, click Stat Γ Z Stats Γ One Sample Γ With Summary.
- Fill in the sample statistics. 4. Under “Perform:”, leave the radio button
at “Hypothesis test for ππ” (the default). 5. Fill in the null hypothesis value and the
alternative hypothesis operator. 6. Click “Compute!”.
To use StatCrunch to conduct a one-sample hypothesis test for ππ using raw data:
- Import/enter the data. 2. For the π‘π‘ distribution, click Stat Γ T Stats Γ One Sample Γ With
Data. For the normal distribution, click Stat Γ Z Stats Γ One Sample Γ With Data.
- Select the appropriate data column. 4. Under “Perform:”, leave the radio button at
“Hypothesis test for ππ” (the default). 5. Fill in the null hypothesis value and the
alternative hypothesis operator. 6. Click “Compute!”.
Example 1 (ππ unknown): The Department of Motor Vehicles office in a certain city claims that its mean wait time for service is at most 14 minutes. A random sample of 70 customers had a mean wait time of 13 minutes with a standard deviation of 3.5 minutes. At significance level πΌπΌ = 0.10, is there enough evidence to reject the office’s claim?
Identify the correct null and alternative hypotheses.
π»π»0: ππ = 14 min (Original claim: ππ β€ 14 min) π»π»π΄π΄: ππ > 14 min
What is the p-value? (Round to three decimal places as needed.) 0.990
State and interpret the appropriate decision for this hypothesis test.
Since the p-value = 0.990 > πΌπΌ = 0.10, we fail to reject π»π»0. There is insufficient evidence to reject the DMV’s claim that its mean wait time for service is at most 14 minutes.
Example 2 (ππ unknown): A research team believes that the mean red blood cell count for adult males (in millions of cells per microliter) is 4.950. The RBC Counts data set contains red blood cell counts for 40 men who took part in the team’s clinical study. Use this data set to test the team’s claim at significance level πΌπΌ = 0.05.
Identify the correct null and alternative hypotheses.
π»π»0: ππ = 4.950 106 cells/dL (Original claim: ππ = 4.950 106 cells/dL) π»π»π΄π΄: ππ β 4.950 106 cells/dL
What is the p-value? (Round to three decimal places as needed.) 0.058
State and interpret the appropriate decision for this hypothesis test.
Since the p-value = 0.058 > πΌπΌ = 0.05, we fail to reject π»π»0. There is insufficient evidence to reject the research team’s claim that the mean red blood cell count for adult males = 4.950 106 cells/dL.
Example 3 (ππ known): A certain drink container has a labeled weight of 12 fluid ounces. The manufacturer claims that the weights of the individual containers have mean ππ = 12.00 fl oz and standard deviation ππ = 0.11 fl oz. Suppose a quality control sample of 36 randomly selected containers has a mean weight of 12.19 fl oz. At significance level πΌπΌ = 0.01, is there enough evidence to reject the manufacturer’s claim?
Identify the correct null and alternative hypotheses.
π»π»0: ππ = 12.00 fl oz (Original claim: ππ = 12.00 fl oz) π»π»π΄π΄: ππ β 12.00 fl oz
What is the p-value? (Round to three decimal places as needed.) 0.000
State and interpret the appropriate decision for this hypothesis test.
Since the p-value = 0.000 < πΌπΌ = 0.01, we reject π»π»0. There is sufficient evidence to reject the manufacturer’s claim that the weights of the individual containers have mean ππ = 12.00 fl oz.
Example 4 (ππ unknown): An environmental activist group claims that the mean pH level of the water in a certain municipal reservoir is less than 6.5, the minimum pH that meets EPA safety standards. In the resulting water quality analysis, 39 randomly selected water samples are found to have a mean pH level of 6.7 with a standard deviation of 0.35. At significance level πΌπΌ = 0.05, is there enough evidence to support the group’s claim?
Identify the correct null and alternative hypotheses.
π»π»0: ππ = 6.5 π»π»π΄π΄: ππ < 6.5 (Original claim: ππ < 6.5)
What is the p-value? (Round to three decimal places as needed.) 1.000
State and interpret the appropriate decision for this hypothesis test.
Since the p-value = 1.000 > πΌπΌ = 0.05, we fail to reject π»π»0. There is insufficient evidence to support the environmental group’s claim that the mean pH level of the water in a certain municipal reservoir is less than 6.5.
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