VI. ASSESSMENT/ACTIVITY Compute the z-value and state whether the z-score locates the raw score x within a sample or within a population 1. x = 50 S = 5 X = 40 2. x = 40 o = 8 u = 52 3. x = 36 S = 6 X = 18 4. x = 74 5. x = 82 S = 10 X = 60 o = 15 u = 75
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- Determine the sample size required to estimate the mean score on a standardized test within 2 points of the true mean with 98% confidence. Assume that s = 15 based on earlier studies. C O A. 1 O B. 18 OC. 407 O D. 306Use the pulse rates in beats per minute (bpm) of a random sample of adult females listed in the data set available below to test the claim that the mean is less than 77 bpm. Use a 0.01 significance level. Click the icon to view the pulse rate data. Pulse Rate Data Pulse Rate (bpm) 38 65 101 66 71 50 78 48 64 36 65 74 39 101 55 40 85 104 61 98 83 72 35 91 76 99 88 75 62 82 36 92 101 89 67 87 102 66 79 90 101 35 58 48 102 91 74 64 74 73 Q - XThe two, box and whisker diagrams below show the percentage marks obtained by two Grade 12 classes. Each class has 26 learners. Class A Class B O 10 20 30 40 50 60 70 80 90 100 The percentage marks for Class A, arranged in order, are given below: F: 58 60 62 62 63 65 65 66 66 66 66 67 69 70 71 73 73 90 [SASAMS database] 75 75 75 H 80 83 85 NOTE: Fis the lowest percentage mark H is a percentage mark between 75 and 80
- Water specimens are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharge is at most 150F, there will be no negative effects on the river's ecosystem. To investigage whether the plant is in compliance with regulations that prohibit a mean discharge water temperature above 150F, researchers will take 50 specimens at randomly selected times and record the temperature of each specimen. The resulting data will be used to test the hypotheses:H0:μ≤150FH0:μ≤150F Ha:μ>150FHa:μ>150F.(a) In the context of this problem, describe Type I and Type II errors.Type I Error:Select an answer A Type I error is not obtaining convincing evidence that the mean water temperature is greater than 150F when in fact it is greater than 150F. A Type I error is obtaining convincing evidence that the mean water temperature is greater than 150F when in fact it is at most 150F. Type II…4Assume that you want to test the claim that the paired sample data come from a population for which the mean difference is Hd = 0. Compute the value of the t test statistic. Round intermediate calculations to four decimal places as needed and final answers to three decimal places as needed. 28 31 20 25 28 27 33 35 y 26 27 26 25 29 32 33 34 X OA. t= -1.185 OB. t= -0.523 OC. t= -1.480 OD. t= -0.690 2 PE 7 2 W Z S x H mmand I E D с 4 1 R F D % L T G 6 B Y 3 I H N U J . 8 I M ( 9 K I O H 2 O L P command Time Remaining: 01:12:18 ; x ( option ? "1 I . Next return
- The manager of a political party wants to find out the association between the campaign spending (Large, medium and small) and the vote obtained (Majority, Average and minority). Use Chi-squared hypothesis test for association to test that there is an association between the variable in campaign spending and the variable in vote obtained with five percent level of significance. The manager uses MS excel to calculate the Chi-squared stat value and the result is 8.874. Critical value: At 1% = 13.277 :At 5% = 9.488 :At 10% = 7.779 1. State the decision rule 2. Conclusion and decision 3. State the alternative hypothesis and the null hypothesisThe manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at random of eight cars and runs the cars until the tires wear out. The data (in kilometers) follow. Is there any difference in mean life of these two brands of tires? Car Brand Brand 36,925 34.318 42,280 1 45.300 36,240 3 4 35.500 32,100 31,950 38,015 37.210 48,360 5 47,800 38,200 37,810 33.500 33,215 6. 7 8. 1. Is this samples independent or paired samples? Explain. 2. Write Ho and H1 3. Calculate Test statistics(T). 4. What is the critical value (T from tables)? 5. What is your conclusions if a = 0.05? 6. Estimate with 90% confidence the different in the means. Which brand would you prefer based on this calculation?Do rats take less time on average than hamsters to travel through a maze? The table below shows the times in seconds that the rats and hamsters took. Rats: 36, 35, 22, 42, 25, 14, 31, 14 Hamsters: 41, 37, 12, 46, 39, 46, 38, 45, 27, 29 Assume that both populations follow a normal distribution. What can be concluded at the a = 0.05 level of significance level of significance? For this study, we should use Select an answer a. The null and alternative hypotheses would be: Ho: Select an answer v Select an answer v Select an answer v (please enter a decimal) H: Select an answer v Select an answer v Select an answer v (Please enter a decimal) b. The test statistic (please show your answer to 3 decimal places.) c. The p-value = (Please show your answer to 4 decimal places.) d. The p-value is ? ♥ a e. Based on this, we should Select an answer f. Thus, the final conclusion is that ... the null hypothesis. O The results are statistically insignificant at a = 0.05, so there is statistically…
- Available below are amounts of arsenic in samples of brown rice from three different regions. The amounts are in micrograms of arsenic and all samples have the same serving size. Use a 0.05 significance level to test the claim that the three samples are from populations with the same mean. Do the amounts of arsenic appear to be different in the different regions? Given that the amounts of arsenic in the samples from region C have the highest mean, can we conclude that brown rice from region C poses the greatest health problem? view the data table of the arsenic amounts. Arsenic Amounts (micrograms) A 4.9 4.9 4.9 5.2 5.3 5.5 5.6 5.7 5.8 5.9 6.1 6.3 B 2.3 3.6 4.5 4.6 4.8 4.8 4.8 5.1 5.1 5.4 5.4 5.5 C 5.7 5.7 6.7 6.9 7.1 7.2 7.2 7.3 7.3 7.4 7.5 7.6 What are the hypotheses to test? H0: mu 1 equals mu 2 equals mu 3μ1=μ2=μ3 H1: At least one…f.A marketing research consultant interested in determining the proportion of customers who use online dating services. A random sample of 2001 people were asked and the count revealed that 15% used online dating services. Data 10 Samples Generate 100 Samples Generate 1000 Samples Reset Plot portion- ht Tail samples - do00 mean- o, 150 std. error=0,008O 0.025 0.950 0.025 0.130 o135 o50 0.16 0.140 0.145 0.155 0.160 0.170 0.175 0.15 0.134 0.166 Compute a 95% confidence interval for the true proportion of people who use online dating. 95% Confidence interval What is the numerical value of the margin of error? If the researcher is using a confidence level of 95% what percent of confidence intervals would not capture the true parameter?
