A sample of 19 elements was used to construct a 99% confidence interval for the population variance, the chi-square value(s) used for this test are: Select one: a. 7.015 to 34.805 O b. 7.633 to 36.191 O c. 6.844 to 38.582 O d. 6.265 to 37.156 ge Next page
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- Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 85 weekly reports showed a sample mean of 18.5 customer contacts per week. The sample standard deviation was 5.4. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel. 90% confidence interval, to 2 decimals: 95% confidence interval, to 2 decimals:Using a 1-7ikert scale, a researcher measured respondents desirability towards cars, and he hypothesized that the respondents rated 4 out of 7 in their desirability towards 2 seat runabout sport electric vehicle. Referring to the SPSS output which of the foliowing statements /are correct? One-Sample Statistics Mean Std. Deviation Std. Error Mean Desirability: 2 Seat Runabout Sport Electric 1000 3.92 1.537 .049 One-Sample Test Test Value 4 95% Confidence Interval of the Mean Difference Sig (2-tailed) Difference Lower Upper Desirability: 2 Seat Runabout Sport Electric -1.626 999 .104 -.079 17 02 The researcher's alternate hypothesis is supported O b All answers are correct Oc The researcher's test value is higher than the average score rated by respondents. O d. The test value falls within the 95% confidence interval of the dataset.Listed in the accompanying table are heights (in.) of mothers and their first daughters. The data pairs are from a journal kept by Francis Galton. Use the listed paired sample data, and assume that the samples are simple random samples and that the differences have a distribution that is approximately normal. Use a 0.05 significance level to test the claim that there is no difference in heights between mothers and their first daughters. ... Question content area top right Part 1 Mother 62.0 65.0 64.7 65.5 65.0 67.0 66.0 66.5 63.0 58.5 Daughter 68.0 69.0 66.5 63.0 68.0 62.0 66.5 66.7 63.5 66.5 Question content area bottom Part 1 In this example, μd is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the daughter's height minus the mother's height. What are the null and alternative hypotheses for the hypothesis test? H0:…
- Below are the SPSS results of a related (dependent) samples t-test. Use the SPSS results to answer Questions 5, 6, and 7. Paired Samples Statistics Mean N Std. Deviation Std. Error Mean Pair 1 Statistics Posttest 57.1250 40 16.73962 2.64677 Statistics Pretest 49.9250 40 15.09014 2.38596 Paired Samples Test Paired Differences t df Sig. (2-tailed) Mean Std. Deviation Std. Error Mean Pair 1 Statistics Posttest - Statistics Pretest 7.200 12.313 ( a ) ( b ) ( c ) .001 Find the mean of the differences ( and standard deviation of the differences (Sd). Replace “a,” “b,” and “c” in the table above with the correct values. Round to the third decimal place. Do you think there is a statistically significant difference between the pretest and posttest at the .05 level of significance?Consider a sample data with x=25 and s=2. Compute a chebyshev’s interval for which at least 89% of the data will lie?A survey of several 9 to 11 year olds recorded the following amounts spent on a trip to the mall: $19.26, $21.87, $15.83, $20.57, $19.31, $24.74, $22.82 Construct the 95% confidence interval for the average amount spent by 9 to 11 year olds on a trip to the mall. Assume the population is approximately normal. Copy Step 2 of 4: Calculate the sample standard deviation for the given sample data. Round your answer to two decimal places.
- Please answer all. Thank you. A national random sample of 20 NCAE Scores from 2010 is listed below. 29,26,13,23,23,25,17,22,17,19 12,26,30,30,18,14,12,26,17,18 Calculate the 95 percent confidence interval for the mean NCAE score if the sample standard deviation is about 5.9407. A.) between 18.22 and 23.48 B.) between 18.07 and 23.63 C.) between negative infinity and 23.15 D.) between negative infinity and 23.05 The value of z selected for constructing a given confidence interval is called A.) confidence B.) error C.) student D.) critical valueA random sample of 120 cars in the drive-thru of a popular fast food restaurant revealed that the mean bill of $29.27 per car. The population standard deviation is $6.17. Find the mean bill for all cars from the drive-thru with 82% confidence.do B AND C
- Given two independent random samples with the following results: n₁ = 16 n₂ = 8 x₁ = 94 X2: $₁ = 16 = 124 $2 = 28 Use this data to find the 90% confidence interval for the true difference between the population means. Assume that the population variances are equal and that the two populations are normally distributed. Copy Data Step 2 of 3: Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.Choose the appropriate statistical test. When computing, be sure to round each answer as indicated. A dentist wonders if depression affects ratings of tooth pain. In the general population, using a scale of 1-10 with higher values indicating more pain, the average pain rating for patients with toothaches is 6.8. A sample of 30 patients that show high levels of depression have an average pain rating of 7.1 (variance 0.8). What should the dentist determine? 1. Calculate the estimated standard error. (round to 3 decimals). [st.error] 2. What is thet-obtained? (round to 3 decimals). 3. What is the t-cv? (exact value) 4. What is your conclusion? Only type "Reject" or Retain"A study of 4 bolts of carpet showed that their average length was 76.2 yards and the sample standard deviation was 5.6 yards. Which of the following is the 80% confidence interval for the mean length per bolt of carpet? O A. 75.8 to 76.6 O B. 72.62 to 79.78 O C. 71.61 to 80.79 D. 75.40 to 77.00
