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1) A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 19
1) A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 19.1 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 18.1, 21.1, 22.1, 23.1, 21.1, 27.1, and 27.1 pounds. What are the degrees of freedom?
| 19.1 | ||
| 7 | ||
| 26.7 | ||
| 6 |
2.A national manufacturer of ball bearings is experimenting with two different processes for producing precision ball bearings. It is important that the diameters be as close as possible to an industry standard. The output from each process is sampled, and the average error from the industry standard is measured in millimeters. The results are presented here.
| Process A | Process B | |
| Sample mean | 2.0 | 3.0 |
| Standard deviation | 1.0 | 0.5 |
| Sample size | 12 | 14 |
The researcher is interested in determining whether there is evidence that the two processes yield different average errors. The population standard deviations are unknown but are assumed equal. What is the null hypothesis?
| H0: µA > µB | ||
| H0: µA ≤ µB | ||
| H0: µA ≠ µB | ||
| H0: µA = µB |
3. Assuming the population variances are known, the variance of thedistribution of differences between twoindependent population means is ________.
| the sum of the two sample sizes for each population | ||
| the sum of the two variances of the two sampling distributions | ||
| the sum of the two means | ||
| the sum of the two standarddeviations of the two sampling distributions |
4.Consider a two-tailed test with a level of confidence of 81.70%. The z-value is _______.
| 2.00 | ||
| 0.89 | ||
| 2.66 | ||
| 1.33 |
5. For a null hypothesis, H0: µ = 4,000, if the 1% level of significance is used and the z-test statistic is +6.00, what is our decision regarding the null hypothesis?
| Reject H1. | ||
| Reject H0. | ||
| Do not reject H0. | ||
| None of these answers apply. |
6. Given the following ANOVA table for three treatments each with six observations:
| Source | Sum of Squares | df | Mean square |
| Treatment | 1,116 | ||
| Error | 1,068 | ||
| Total | 2,184 |
What is the treatment mean square?
| 534 | ||
| 558 | ||
| 71.2 | ||
| 71.4 |
7. Given the following ANOVA table for three treatments each with six observations:
| Source | Sum of Squares | df | Mean square |
| Treatment | 1,116 | ||
| Error | 1,068 | ||
| Total | 2,184 |
What is the computed value of F?
| 7.48 | ||
| 8.84 | ||
| 7.84 | ||
| 8.48 |
8. If the alternate hypothesis states that µ ≠ 4,000, where is the rejection region for the hypothesis test?
| In both tails | ||
| In the lower or left tail | ||
| In the center | ||
| In the upper or right tail |
9.i f the null hypothesis that two means are equal is true, where will 97% of the computed z-values lie between?
| ±2.17 | ||
| ±2.33 | ||
| ±2.07 | ||
| ±2.58 |
10. In an ANOVA problem involving three treatments and 19 observations per treatment, SSE = 648. The MSE for this scenario is _______.
| 12 | ||
| 54 | ||
| 29.5 | ||
| 216 |
Expert Solution
1) 6
2) HO : μ1 = μ2
HA : μ1 ≠ μ2
3) The sum of the two variances of the two sampling distributions
4) 1.33
5) Reject the Ho.
6) 558
7) 7.84
8) Both tails
9) ± 2.17
10) 12
Step-by-step explanation
1) We have here,
Sample size =n=7
Degree of freedom =n-1 =7-1 = 6
Degree of freedom =6
2) The claim is to test whether there is evidence that the two processes yield different average errors.
Therefore, the hypothesis can be written as
HO : μ1 = μ2
HA : μ1 ≠ μ2
3) Assuming the population variances are known, the population variance of the difference between two means is the sum of the two population variances. The variance of the sampling distribution of the difference between means is equal to the sum of the variance of the sampling distribution of the mean for the two populations.
4) At 80.30% confidence level the z is ,
= 1 - 81.70% = 1 - 0.8170 = 0.183
α/2 = 0.0915
Zα/2 = Z0.0915 = 1.33 ( Using z table )
5) Given test statistic z= 6
P(z > 6 ) = 1 - P(z < 6 ) =1 - 1=0 using standard normal distribution table
The p-value is 0
The level of significance is 1% which is 0.01
Decision: Since the p-value(0) is less than the significance level(0.01), the we reject the null hypothesis.
Conclusion: There is insufficient evidence to reject the claim that the mean is 4000.
6) The treatment mean square = Sum of Squares treatment / degrees of freedom of treatment
=1116/(3 - 1)
=558
7)
| Source | Sum of Squares | df | Mean square |
| Treatment | 1,116 | 2 | 558 |
| Error | 1,068 | 15 | 71.2 |
| Total | 2,184 | 17 |
The value of F computed is
= MSTR/MSE
= 558/71.2
= 7.837
= 7.84
8) This is the two tailed test,
The null and alternative hypothesis is ,
H0 : μ = 4000
Ha : μ ≠ 4000
The rejection region for the hypothesis test in both tails
z obtained < z critical or z obtained > z critical
9) We are given that: P(-z < Z < z) = 0.97
This means that P(Z < z) - P(Z < -z) = 0.97
2 * P(Z < z) - 1 = 0.97.
2 * P(Z < z) = 1.97
P(Z < z) = 0.985
From that standard normal table,
P(Z < 2.17) = 0.985
Therefore, the value of z is
z ± 2.17
10) We have 3 treatments and 21 observations per treatment. Therefore, the total number of observation (n) is = 3 * 19 = 57 and The total degrees of freedom (TSS) = n - 1 = 57 - 1 = 56 and the degrees of freedom for treatment (SST) is = 3 - 1 = 2
Thus the degrees of freedom for error ( SSE) = 56 - 2 = 54
Since we know that
Mean sum of square of error (MSE) = Sum of square due to error (SSE) / degrees of freedom for error
= 648 / 54 = 12
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