Hypothesis Test Results:- **μ**: Mean of variable- **Null Hypothesis (H₀)**: μ = 2.2- **Alternative Hypothesis (Hₐ)**: μ > 2.2| Variable | Sample Mean | Std. Err. | DF | T-Stat | P-value ||----------|-------------|-----------|----|--------|---------|| Length | 2.46821 | 0.276896 | 399| 0.968631| 0.1667 |**Explanation:**- **Sample Mean (2.46821)**: The average measured value from the sample data.- **Standard Error (0.276896)**: The estimated standard deviation of the sample mean, indicating variability.- **Degrees of Freedom (DF = 399)**: The number of independent values in the dataset.- **T-Statistic (0.968631)**: The test statistic used to determine how far the sample mean is from the null hypothesis mean.- **P-value (0.1667)**: The probability of observing the data, or something more extreme, if the null hypothesis is true. A lower P-value suggests stronger evidence against the null hypothesis.

The random variable X follows normal distribution. The given output represents one-sample t-test…

Answered: Hypothesis test results: H: Mean of…

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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rse ations Suppose a study was done to determine whether the average amount of sleep that students get the night before an exam is less than 6 hours. An SRS of 100 students from a university was taken, and a mean of 5.5 hours was computed from the sample. The figure below provides the results of the analysis. Normal calculator Density 04 0.3 02 0.1 0 -3 -2 Mean: 0 Std. Dev. 1 Prob(X<= -2.5 )=0.0062096654 Snapshot Close Compute What is your conclusion based on the results of this study? (Assume a significance level of 0.05.) O On average, students typically get less than 6 hours of sleep the night before an exam. O On average, students typically get 6 hours of sleep the night before an exam. O On average, students typically get more than 6 hours of sleep the night before an exam. O On average, students typically get 5.5 hours of sleep the night before an exam.
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Summary of Fit RSquare 0.673175 0.659743 RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) 11.36538 106.8831 77 Analysis of Variance Sum of Source DF Squares Mean Square F Ratio Model 3 19422.399 6474.13 50.1203 Error 73 9429.549 129.17 Prob > F C. Total 76 28851.948 |t| Intercept 29.593628 7.186991 4.12 <.0001* Sugars 3.8623817 0.348158 11.09 <.0001* Sodium 0.0089653 0.016746 0.54 0.5940 Carbo 3.3058183 0.408135 8.10 <.0001* 1. Using the above JMP output, which of the predictors present concern to the overall model? Assume the response, y, is calories found in a cup of cereal.
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Estimated Speed Smashed into hit n = 16 n = 16 M = 40 M = 34 SS = 510 SS = 414 Use a one-tailed test with α = .01. The t-statistic is
Nonparametric: A small study is conducted to compare birth weight in babies born to mothers who do not smoke to those that smoke more than ½ pack/day. Is there a significant difference between birth weights due to maternal smoking status? Use the Mann-Whitney U test with a 5% level of significance. What is U1 (non-smokers)? Birth weights of infants in two groups of mothers. (Observed data) Nonsmokers Heavy Smokers (≥1/2 pack/day) 8.6 7.0 8.5 5.2 6.3 6.1 9.3 6.7 8.0 6.6 6.8 7.4 4.9 6.3
Testing:H0:μ=38.5H1:μ<38.5Your sample consists of 35 subjects, with a mean of 38.3 and a standard deviation of 1.93.Calculate the test statistic, rounded to 2 decimal places. t =
Testing:H0:μ=4.98H1:μ>4.98Your sample consists of 35 subjects, with a mean of 7 and standard deviation of 4.44.Calculate the test statistic, rounded to 2 decimal places. t=
Chicken diet and weight: Compare the effects of six types of feed: casein, horsebean, linseed, meat meal, soybean, and sunflower. The ANOVA output below can be used to test for differences between the mean weights of chicks on different diets. df Sum Sq Mean Sq F value Pr(>F) feed 5 231129.16 46225.83 15.36 0.0000 residuals 65 195556.02 3008.55 Conduct a hypothesis test to determine if these data provide convincing evidence that the average weight of chicks varies across some (or all) groups. Figures and summary statistics are shown below. Test to see if there is a difference between mean weight of chicks between at least two of the diets using overall a = 0.05. 400 350- 300 - 250 200 150 - 100 casein horsebean linseed meatmeal soybean sunflower Mean SD casein 323.58 64.43 12 horsebean 160.2 38.63 10 linseed 218.75 52.24 12 meatmeal 276.91 64.9 11 soybean 246.43 54.13 14 sunflower 328.92 48.84 12 Weight (in grams)
Please find Percentile ranking
Assume a media agency reports that it takes television streaming service subscribers in the U.S. an average of 6.60 days to watch the entire first season of a television series, with a standard deviation of 4.21 days. Scarlett is an analyst for an online television and movie streaming service company that targets to the 18-50 age bracket. She wants to determine if her company's customers exhibit shorter viewing rates for their series offerings. Scarlett plans to conduct a one-sample z-test, with significance level of a = 0.05, to test the null hypothesis, Ho: µ = 6.60 days, against the alternative hypothesis, H1: µ < 6.60 days. The variable u is the mean amount of time, in days, that it takes for customers to watch the first season of a television series. Scarlett selects a simple random sample of 740 customers who watched the first season of a television series from the company database of over 25,000 customers that qualified. She compiles the summary statistics shown in the table.

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Hypothesis test results: H: Mean of variable Ho : µ= 2.2 HA : H> 2.2 Variable Sample Mean Std. Err. Length DF T-Stat P-value 2.46821 0.276896 399 0.968631 0.1667