The Population Standard Deviation - Assignment Example

Stat 2300 Summer Exam 2 _________________________ Location_______________________ μ0 = $130 per trip, n = 45, =$135.25 per trip , σ = $10.50 and α = .05 The null and alternate hypotheses are H0: μ ≤ $130 H1: μ > $130 (Right-tailed test) The population standard deviation, σ is known, therefore, the selected test is z-test for a Population Mean. The test statistic is The p-value is p-value (right-tailed) = .0004 {Using Excel Function =1-NORMSDIST(3.354)} Decision: Reject H0, as p-value = .0004 is less than α = .05. In conclusion, there is enough evidence to suggest that the average amount spent at the competitor is higher than at Costco. 2. μ0 = 12 ounces, n = 20, =11.80 ounces, s = 0.8 and α = .01 The null and alternate hypotheses are H0: μ = 12 H1: μ ≠ 12 (Two-tailed test) The population standard deviation, σ is unknown, therefore, the selected test is t-test for a Population Mean. The degrees of freedom are df = n – 1 = 20 – 1= 19 The test statistic is The p-value is p-value (two-tailed) = .2775 {Using Excel Function =TDIST(ABS(-1.118),19,2)} Decision: Fail to reject H0, as p-value = .2775 is greater than α = .01. In conclusion, there is not enough evidence to suggest that the filling weight is significantly different from 12 ounces. It is recommended to the brewery that they do nothing to analyze the accuracy of the bottling process, as the bottling process is in control. 3. p0 = .82, n = 200 and x =150 and sample proportion, The null and alternate hypotheses are H0: p ≥ .82 H0: p < .82 (Left-tailed test) Selecting the significance level, α = .05 (Not given). The selected test is z-test for a Population Proportion. The test statistic is The p-value is p-value (left-tailed) = .0050 {Using Excel Function =NORMSDIST(-2.577)} Decision: Reject H0, as p-value = .0050 is less than α = .05. In conclusion, there is strong evidence (p-value < .01) to suggest that the proportion of teenage drivers in Massachusetts who use a cell phone while driving is less than 82%. 4. σ0 = $300, n = 15, s =$332.45 and α = 0.10 The null and alternate hypotheses are H0: σ2 ≤ 90000 H1: σ2 > 90000 (Right-tailed test) The selected test is Chi-square Variance Test. The degrees of freedom are df = n – 1 = 15 – 1= 14 The test statistic is The p-value is p-value (right-tailed) = .2461 {Using Excel Function ==CHIDIST(17.192,14)} Decision: Fail to reject H0, as p-value = .2461 is greater than α = .10. In conclusion, there is not enough evidence to suggest that the standard deviation is more than $300. Thus, there is no concern about the consistency of business. 5. μ0 = 58.7 years, α = .05 and H0: μ ≤ 58.7 a. Type II error in the context of this problem: Not rejecting the null hypothesis H0: μ ≤ 58.7 when it is false. In other words, suggesting that the average retirement age for French citizen is not more than 58.7 years, in fact it is more than 58.7 years. The impact of making a Type II error is to concluding that the new bill has not raised the average age at which people actually retire. b. σ(assumed) = 5 years, n = 40 and μa = 60.5 years The selected test is one-sided test, therefore, z* = zα = z.05 = 1.645 β = area under the normal curve to the left of β = area under the normal curve to the left of β = 0 .2637 {Using Excel Function =NORMSDIST(-0.632)} The probability of committing a Type II error if the actual retirement age is 60.5 years old is about 0.2637. c. same condition and μa = 61.5 years β = area under the normal curve to the left of β = area under the normal curve to the left of β = 0.0289 {Using Excel Function =NORMSDIST(-1.897)} The probability of committing a Type II error if the actual retirement age is 61.5 years old is about 0.0289. d. β = 0.10 Sample size determination to achieve specified values of α and β Here, zβ is the point on the z-curve that gives a right-hand tail area equal to β. Therefore, zβ = z1-0.10 = z.90 = 1.282 = 67 (rounded up) A sample size of 67 (minimum) should be used if the group is willing to accept a 0.10 probability of making a Type II error when the actual average retirement age is μ = 60.5. 6. First section: = 77.4, s1 = 10.8 and n1 = 18 Second section: = 74.1, s1 = 12.2 and n1 = 14 a. α = .05 The null and alternate hypotheses are (Two-tailed test) The selected test is F-test for Equality of Variance. The degrees of freedom are Numerator, df1 = n1 – 1 = 18 – 1= 17 Denominator, df2 = n2 – 1 = 14 – 1= 13 The test statistic is The p-value is p-value (two-tailed) = .6274 {Using Excel Function =2*(1-FDIST(0.784,17,13))} Decision: Fail to reject H0, as p-value = .6274 is greater than α = .05. In conclusion, there is not enough evidence to suggest that there is a difference in the variance between the scores of the two groups. b. α = .05 The null and alternate hypotheses are (Two-tailed test) The selected test is t-Test for Difference in Means Assuming Equal Variances. This is because the results of hypothesis test in part a suggested that there is no difference in the variance between the scores of the two groups. The degrees of freedom are df = n1 + n2 – 2 = 18 + 14 – 2 = 30 The pooled variance is The test statistic is The p-value is p-value (two-tailed) = .4241 {Using Excel Function =TDIST(ABS(0.810),30,2)} Decision: Fail to reject H0, as p-value = .4241 is greater than α = .05. In conclusion, there is not enough evidence to suggest that there is a difference in the mean between the scores of the two groups. 7. The sample mean and sample standard deviation for difference in weight for participants before and after the diet program are Weight Before Weight After d = Before – After d2 158 151 7 49 205 200 5 25 170 169 1 1 189 179 10 100 149 144 5 25 135 129 6 36 = 34 = 236 D0 = 5 pounds and α = 0.05 The null and alternate hypotheses are H0: μd ≤ 5 H1: μd > 5 (Right-tailed test) The selected test is Paired t-Test. The degrees of freedom are df = n – 1 = 6 – 1= 5 The test statistic is The p-value is p-value (right-tailed) = .3015 {Using Excel Function =TDIST(ABS(0.555),5,1)} Decision: Fail to reject H0, as p-value = .3015 is greater than α = 0.05. In conclusion, there is not enough evidence to suggest that the participants in diet center’s weight loss program lose more than 5 pounds within a month. Therefore, the diet center’s claim that it has the most effective weight loss program in the region is not correct. 8. Recent: x1 = 67 and n1 = 150 Three years ago: x2 = 58 and n2 = 140 α = 0.01 An estimate of is The null and alternate hypotheses are (Right-tailed test) The selected test is Large sample z-Test for Difference in Proportions. The test statistic is The p-value is p-value (right-tailed) = .2890 {Using Excel Function =1-NORMSDIST(0.556)} Decision: Fail to reject H0, as p-value = .2890 is greater than α = 0.01. In conclusion, there is not sufficient evidence to suggest that more people are now using LinkedIn to search for jobs as compared to three years ago. Read More