Introduction to Statistics Exam - Assignment Example

MATH 215 An Introduction to Statistics Final Examination Version B Question a. x x2 20 400 30 900 25 625 15 225 25 625 = 115 = 2775 Sample mean,Sample standard deviation, df = n – 1 = 5 – 1 = 4 For 80% confidence with 4 degrees of freedom the critical value of t is 1.533. Therefore, an 80% confidence interval for the population mean use (μ) per week is = 19.09 to 26.91 An 80% confidence interval for the population mean use (μ) per week is between 19.09 and 26.91. b. For 80% confidence with 4 degrees of freedom the critical value of χ2 are 7.779 and 1.064. Therefore, an 80% confidence interval for the population variance (σ2) of glove use per week is An 80% confidence interval for the population standard deviation (σ) of glove use per week is An 80% confidence interval for the population standard deviation (σ) of glove use per week is between 4.09 and 11.05. Question 2 For 85% confidence, z = 1.440 = 1867 (rounded up) A sample of at least 1867 bottled water drink should be selected to estimate the population proportion that violates at least one government standard to within a margin of error of 0.01 with 85% confidence. Question 3 = $850 per month, = $909.25 per month, n = 64 and σ = $200 per month The null and alternate hypotheses are (The mean apartment rental now does not exceed $850 per month.) (The mean apartment rental now exceeds $850 per month.) The selected level of significance, α is 0.025. The normal distribution will be used for the test, as population standard deviation is known. The test statistic is The p-value is p-value (right-tailed) = 1 – 0.9911 = 0.0089 Decision: Reject the null hypothesis H0, as p-value = 0.0089 < α = 0.025. Conclusion: There is sufficient evidence that the mean apartment rental now exceeds $850 per month. Question 4 Men: n1 = 400 and x1 = 260 Women: n2 = 500 and x2 = 350 The null and alternate hypotheses are (There is no significant difference in the proportions of men and women.) (There is a significant difference in the proportions of men and women.) The selected level of significance, α is 0.01. The two sample proportions are Then, and The value of , , , and are Because each of , , , and values is greater than 5, both sample sizes are large. Consequently, we can use the normal distribution to test the difference between . The equal to (=) sign in the alternative hypothesis indicates that the test is two-tailed test. From the normal distribution table, for a 0.01 significance level, the critical values of z for two-tailed test are -2.576 and 2.576. Therefore, decision rule will be Reject H0 if z > 2.576 or z < -2.576 Otherwise, do not reject H0. The pooled sample proportion is and The estimate of the standard deviation of is The value of the test statistic z for is Decision: Fail to reject the null hypothesis H0, as z = -1.60 falls in-between -2.576 and 2.576 that is in the non-rejection region. Conclusion: There is insufficient evidence that there is a significant difference in the proportions of men and women who rate slot machines as their favourite game. Thus, At the 1% significant level, it cannot be concluded there is a statistically significant difference in the proportions of men and women who rate slot machines as their favourite game. Question 5 Before After d = After - before d2 98 90 -8 64 83 79 -4 16 74 70 -4 16 92 89 -3 9 74 71 -3 9 62 54 -8 64 = -30 = 178 Sample mean for paired differences: Sample standard deviation for paired difference: The null and alternate hypotheses are (The jogging program did not reduce resting pulse rate) (The jogging program reduced resting pulse rate.) The selected level of significance, α is 0.01. The sample size is small (n < 30) and is unknown, therefore, using t distribution for the test. The degrees of freedom are df = n – 1 = 6 – 1 = 5 At level of significance of 1% with 5 degrees of freedom, the critical value of t for a left-tailed test is -3.365. Therefore, decision rule will be Reject H0 if t < -3.365. Otherwise, do not reject H0. The test statistic is Decision: Reject the null hypothesis H0, as t = -5.175 < 1.415. Conclusion: At the 1% significant level, there is sufficient evidence that the two-month jogging program reduced resting pulse rate of women aged 40 to 50. Question 6 The null and alternate hypotheses are The distribution of colours of plain chocolate candies in a box is not the same as hypothesized. The selected level of significance, α is 0.025. There are four categories, therefore using the Chi-square distribution for the test. The degrees of freedom is df = k – 1 = 4 – 1 = 3 At level of significance of 2.5% with 3 degrees of freedom, the critical value of the χ2 is 9.348. Therefore, decision rule will be Reject H0 if χ2 > 9.348. Otherwise, do not reject H0. The test statistic is (See calculation in the table) Colour Observed Frequency O Proportion p Expected Frequency E = np O – E (O – E)2 (O – E)2/E Brown 180 0.40 200 -20 400 2.000 Yellow 170 0.30 150 20 400 2.667 Red 80 0.15 75 5 25 0.333 Blue 70 0.15 75 -5 25 0.333   n = 500 Sum = 5.333 Decision: Fail to reject the null hypothesis H0, as the value of the test statistic χ2 = 5.333 is less than the critical value of χ2 = 9.348. Conclusion: No, the observed data does not contradict the company’s claim. Question 7 The null and alternate hypotheses are (All three population means are equal.) Not all three population means are equal. The selected level of significance, α is 0.05. The selected test is Analysis of Variance (ANOVA). The degrees of freedom are Degrees of freedom for the numerator, df1 = k – 1 = 3 – 1 = 2 Degrees of freedom for the denominator, df2 = n – k = 12 – 3 = 9 At level of significance of 5% with 2 and 9 degrees of freedom, the critical value of the F is 4.26. Therefore, decision rule will be Reject H0 if F > 4.26. Otherwise, do not reject H0. SSB = 20.6667 and SSW = 13 (Given) SST = SSB + SSW = 20.6667 + 13 = 33.6667 The variance between samples (MSB) and the variance within samples (MSW) are MSB = SSB/(k – 1) = 20.6667/2 = 10.3333 MSW = SSW/(n – k) = 13/9 = 1.4444 Below ANOVA table summarizes the above calculations. ANOVA Source of Variation df SS MS F Fcrit Between Groups 2 20.6667 10.3333 7.154 4.26 Within Groups 9 13 1.4444 Total 11 33.6667       The value of the test statistic F is Decision: Reject the null hypothesis H0, as the value of the test statistic F = 7.154 is greater than the critical value of F = 4.26. Conclusion: There is insufficient evidence that there is no difference in the mean weekly hours that executives in the three industries spend at their desktop computers. In other words, there is a difference in the mean weekly hours that executives in the three industries spend at their desktop computers. Question 8 a. The type of relationship that would be expected between temperature and water consumption is positive that is as daily temperature increases in the city, the consumption of water also increases and vice-versa. b. Scatter diagram c. Least squares regression line with water consumption (y) as the dependent variable, and temperature (x) as the independent variable: Thus, least squares regression line is given by: d. Scatter diagram with regression line For x = 10, For x = 30, e. The null and alternate hypotheses are H0: B = 0 (The slope is zero) H1: B > 0 (The slope is positive) The selected level of significance, α is 0.025. Since, n = 7 < 30 and is unknown, therefore, using the t distribution to make the test about B. Area in the right tail = α = 0.025 df = n – 2 = 7 – 2 = 5 At level of significance of 0.025 with 5 degrees of freedom, the right-tail critical value of t is 2.571. Therefore, decision rule will be Reject H0 if t > 2.571. Otherwise, do not reject H0. The test statistic is Decision: Reject the null hypothesis H0, as the value of the test statistic t = 7.2545 is greater than the critical value of t = 2.571. Thus, it can be concluded that the slope of the regression line is positive. f. For x = , Area in each tail = α/2 = 0.5 – (0.90/2) = .05 df = n – 2 = 7 – 2 = 5 t = 2.015 Hence, the 90% prediction interval for for x =is A 90% prediction interval for the water consumption on a particular day when the temperature is is between and . Read More