**Title: Relating Confidence Intervals to Hypothesis Testing (One Sample)****Course: Business Data Analytics****Section: 4.2.2 - Confidence Intervals for Population Means**---### Challenge Activity#### OverviewAn auto maker is interested in gathering data about the lifespan of its transmissions. The transmissions are run continuously, and the number of miles they operate before a failure is recorded. The auto maker claims that its transmissions can run continuously for over 150,000 miles before a failure occurs. The following table presents data from a sample study conducted to verify this claim. ---#### Dataset and Calculations- **Miles (1000s of Miles):** 151.2- **Mean:** 151.2- **Variance:** 19.22- **Observations:** 44- **Hypothesized Mean:** 150- **Degrees of Freedom (df):** 43- **t Stat:** 1.889- **P(T<=t) One-Tail:** 0.0338- **t Critical One-Tail:** 1.681- **P(T<=t) Two-Tail:** 0.0676- **t Critical Two-Tail:** 2.017- **Confidence Level (95.0%):** 1.129#### Explanation of the GraphThe graph represents a two-tailed t-distribution, displaying the t critical values and the sample t-statistic. - **X-Axis (t-values):** Ranges from -3 to 3 indicating the critical regions for the one and two-tailed tests.- **P-value Box:** Shows the computed p-value of 0.0676 for a two-tailed test and 0.0338 for a one-tailed test.- **t-value for Sample:** 1.234- The blue curve depicts the t-distribution, where shaded regions correspond to critical values and overall confidence intervals.#### Formulas and Placeholders- $ n = \text{Ex: 5} $- $ \bar{x} = \text{Ex: 1.2} $- $ s = \text{Ex: 5} $- Degrees of Freedom = Ex: 5- $ s = \text{Ex: 1.234} $---#### Instructions for Students1. Review the provided

An auto maker is interested in information about how long transmissions last. A sample of…

Answered: An auto maker is interested in…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A middle school teacher believes that a reading rewards program results in a higher number of pages read per year after the program is completed. The teacher chooses a simple random sample of students and records the number of pages read per year before starting the program and after the program is completed. The results of the study are below. What are the population parameters? Pick Mean Variance Observations Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail Confidence Level Before 813.25 192250.25 182875.22 100 100 0 198 1.497 0.068 1.653 0.136 1.972 95% After 904.95 What is the level of significance? Ex: 0.12 Is the two-sample hypothesis test a paired or unpaired t-test? Pick The null hypothesis is Ho: after Pick The alternative hypothesis is Ha : Mafter Pick Should Ho be rejected or does Ho fail to be rejected? Pick What conclusion can be drawn from the data? Pick Mbefore before ✓evidence exists to support the claim…
Assume the samples are random and independent, the populations are nomally distributed, and the population variances are equal. The table available below shows the prices (in dollars) for a sample of automobile batteries. The prices are classified according to battery type. At a = 0.10, is there enough evidence conclude that at least one mean battery price is different from the others? Complete parts (a) through (e) below. E Click the icon to view the battery cost data. (a) Let u1. P2. H3 represent the mean prices for the group size 35, 65, and 24/24F respectively. Identify the claim and state Ho and H. H Cost of batteries by type The claim is the V hypothesis. Group size 35 Group size 65 Group size 24/24F 101 111 121 124 D 146 173 182 278 124 140 141 89 (b) Find the critical value, Fo, and identify the rejection region. 90 79 84 The rejection region is F Fo, where Fo = (Round to two decimal places as needed.) (c) Find the test statistic F. Print Done F= (Round to two decimal places as…
Variability within groups can have a huge impact on your analyses. With increased variability, you are ________ likely to find significant differences between the groups. A. More B. Less
You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. A sample of 24 night students and 28 day students and their GPA's are recorded below. Test the claim using a 1% level of significance. Assume the population variances are unequal and that GPAs are normally distributed. Give answers to 4 decimal places. GPA-Night GPA-Day 3.05 3.38 2.91 3.02 3.27 3.4 3.29 2.73 3.3 3.24 3.35 2.69 3.39 3.25 3.52 3.33 3.39 2.33 3.25 3.51 3.08 3.07 3.27 2.9 3.32 3.48 3 2.94 3.62 2.93 3.64 3.31 3.54 2.62 3.45 2.66 3.29 3.27 3.56 2.76 3.15 2.7 3.4 3.27 3.51 2.96 3.54 2.86 3.24 2.72 2.64 3.14 What are the correct hypotheses? Note this may view better in full screen mode. Select the correct symbols in the order they appear in the problem.H0: Select an answer x̅₂ s² x̅₁ σ² μ₂ μ₁ p ? = ≠ > < ≥ ≤ Select an answer 0 μ₂ x̅₂ x̅₁ σ² p s² μ₁ H1: Select an answer σ² x̅₂ x̅₁ μ₁ μ₂…
_________ are used to display actual data points for individual-level data. Would this answer be Univariate Statistics?
Find the mean, the median, and all modes for the data in the given frequency distribution. (Round your answers to one decimal place. If there is more than one mode, enter your answer as a comma-separated list. If an answer does not exist, enter DNE.) mean median mode(s) Points Scored by Lynn Points scored in abasketball game Frequency 2 9 6 8 7 9 10 7 11 2 13 4 14 2
Box plots (Excel)– Analyze each of these cycle time data but on the same chart: C7, C8, C9. That means one chart showing three different box plots, with a single conclusion that addresses all three. Interpret outside value, mean score, N spread, Fence, upper lower and upper outer fences, adjacent, and outside and far out value. C7 C8 C9 19.98211 20.7142 20.21992 20.01223 20.65907 20.00799 20.02408 20.6437 20.02974 19.94534 20.45958 19.94026 19.86814 20.63716 19.796 20.18784 20.65365 20.01327 19.92074 20.54906 20.01064 20.0778 20.5228 20.07752 19.88045 20.47224 19.95005 20.08633 20.61675 20.13076 19.99881 20.62622 20.06755 20.04184 20.42231 20.13078 19.7936 20.61609 19.88701 19.93159 20.52705 19.87559 20.19998 20.58901 19.88053 20.18833 20.49057 19.9302 20.12368 20.68086 19.9738 19.91343 20.57999 19.8936 20.1602 20.34386 20.02487 20.09396 20.62949 20.09329 20.11419 20.66649 20.00717 19.9063 20.71718 20.04545 20.12964 20.43237 19.95537…
Consider the following data: −10, −10, 0, 3, −10, 3, 3 Step 1 of 3: Calculate the value of the sample Variance. Round your answer to one decimal place. Step 2 of 3: Calculate the value of the sample deviation. Round your answer to one decimal place. Step 3 of 3: Calculate the value of the range.
Find the mean, the median, and all modes for the data in the given frequency distribution. (Round your answers to one decimal place. If there is more than one mode, enter your answer as a comma-separated list. If an answer does not exist, enter DNE.) mean median mode(s) Points Scored by Lynn Points scored in abasketball game Frequency 2 8 8 5 9 8 13 4 14 1 18 3 21 1
Consider the following sample data. Sample A: 11, 22, 33 Sample B: 81, 92, 103 Sample C: 1,100; 1,111; 1,122 (a) Find the mean and standard deviation for each sample. Sample A: Sample B: Sample C: Mean Sample Standard Deviation (b) What does this exercise show about the standard deviation? multiple choice The idea is to illustrate that the standard deviation is not a function of the value of the mean. The idea is to illustrate that the standard deviation is a function of the value of the mean.
A magazine claims that the mean amount spent by a customer at Burger Stop is greater than the mean amount spent by a customer at Fry World. The results for samples of customer transactions for the two fast food restaurants are shown below. At x = 0.01, can you support the magazine's claim? Assume the population variances are equal. Assume the samples are random and independent, and the populations are normally distributed. Complete parts (a) through (e) below. Burger Stop Fry World X₁ = $11.22 X₂ = $10.59 $₁ = $0.79 $2 = $0.64 n₂ = 10 n₁ = 12 (a) Identify the claim and state Ho and H₂. Which is the correct claim below? O A. "The mean amount spent by a customer at Burger Stop is greater than the mean amount spent by a customer at Fry World." OB. "The mean amount spent by a customer at Burger Stop is equal to the mean amount spent by a customer at Fry World." O c. "The mean amount spent by a customer at Burger Stop is not equal to the mean amount spent by a customer at Fry World." O D.…

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