Preliminary data analysis indicate that it is reasonable to use a t-test to carry out the specified hypothesis test. Perform the t-test using the critical-valueapproach.Use a significance level of a=0.05 to test the claim that u = 35.8. The sample data consists of 10 scores for which x =44.2 and s = 7.3.Ho : u+35.8O:u= 35.8Test statistic: t= 3.64. Critical values: t =±2.262. Reject H,: u= 35.8.There is sufficient evidence to support the claim that the mean is different from 35.8.Ho : u= 44.2H:u=44.2Test statistic: t= 3.64. Critical values: t= ±2.262. Reject H: u= 44.2.There is sufficient evidence to support the claim that the mean is different from 44.2.H: u=44.2H: u=44.2Test statistic:1= 3.64. Critical values: t = ±2.262. Reject H: u = 44.2.There is sufficient evidence to support the claim that the mean is different from 44.2.H:u= 35.8Hu= 35.8Test statistic:t=3.64. Critical values: 1= ±2.262. Reject H: u= 35.8.There is sufficient evidence to support the claim that the mean is different from 35.8.

Hypotheses: H0:μ=35.8H1:μ≠35.8

Answered: Preliminary data analysis indicate that…

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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identify the test statistic. Identify the P-value.
Big babies: The National Health Statistics Reports described a study in which a sample of 310 one-year-old baby boys were weighed. Their mean weight was 24.4 pounds with standard deviation 5.3 pounds. A pediatrician claims that the mean weight of one-year-old boys is greater than 25 pounds. Do the data provide convincing evidence that the pediatrician's claim is true? Use the α = 0.10 level of significance and the P-value method with the TI-84 Plus calculator. Compute the P-value. Round the answer to at least four decimal places. P-value = ____ How do I solve this problem using the TI-84 calculator?
Using the results obtained for a study on website development below, how would we interpret the results of this hypothesis test? The claim is the percent of customers who abandon a web page due to long load times is different from 34%. The significance level is 0.10. The test statistic is 2.50. Select the correct answer below: Reject H0. There is strong evidence to conclude that the percent of customers who abandon a web page due to long load times is greater than 34%. Reject H0. There is NOT strong evidence to conclude that the percentage of customers who abandon a web page due to long load times is greater than 34%. Fail to reject H0. There is strong evidence to conclude that the percent of customers who abandon a web page due to long load times is greater than 34%. Fail to reject H0. There is NOT strong evidence to conclude that the percentage of customers who abandon a web page due to long load times is greater than 34%.
A study was done using a treatment group and a placebo group. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.05 significance level for both parts. Treatment Placebo μ μ1 μ2 n 27 39 x 2.38 2.65 s 0.87 0.61 a. Test the claim that the two samples are from populations with the same mean. What are the null and alternative hypotheses? A. H0: μ1≠μ2 H1: μ1<μ2 B. H0: μ1<μ2 H1: μ1≥μ2 C. H0: μ1=μ2 H1: μ1>μ2 D. H0: μ1=μ2 H1: μ1≠μ2 Your answer is correct. The test statistic, t, is (Round to two decimal places as needed.)
Do cars really get better mileage per gallon on the highway? The table shows results from a study of the MPG (miles per gallon) of cars both in the city and on the highway. Assume that the two samples are randomly selected, independent, the population standard deviations are not know and not considered equal. At the 0.1 significance level, test the claim that the mpg on the highway is better than in the city. MPG on the Highway 35.6 34.3 32.2 33.9 31.1 27.1 33.3 33.4 29.3 33.5 31.4 33.2 33.5 30.8 33.8 MPG in the City 26.4 25.3 18.6 25.6 24.7 24.6 25.1 22.4 29.3 23.7 23.4 22 24 23.6 25.5 What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.)H0: Select an answer p x̄₁ p₁ σ₁² μ₁ μ₂ μ μ(Highway) x̄₂ p̂₁ s₁² p₂ ? ≤ ≠ < ≥ = > Select an answer p₁ p₂ p̂₁ μ(City) μ μ₁ μ₂ x̄₁ x̄₂ s₁² σ₁² p H1: Select an answer p₂ μ(Highway) p̂₂ σ₂² x̄₁ x̄₂ s₂² μ₁ μ₂ μ p₁ p ? < ≠ = ≥ ≤ > Select an answer p₂ p₁ μ₁ σ₁²…
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"
please help me dtermine the test statistic AND P-value?
You are conducting a study to see if the proportion of men over 50 who regularly have their prostate examined is significantly more than 0.52. You use a significance level of a = 0.05. Ho:p = 0.52 H1:p> 0.52 You obtain a sample of size n = 556 in which there are 308 successes. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is... less than (or equal to) a greater than a This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that. There is sufficient evidence to warrant rejection of the claim that the proportion of men over 50 who regularly have their prostate examined is more than 0.52. There is not sufficient evidence to warrant rejection of the claim that the proportion of men over 50 who regularly have their prostate examined is more…
Salma claimed that the gas price in Blue Park neighborhood (mean of $3.65) is higher than that in Green Spring (mean of $3.59). She obtained the t-value of 2.223 after calculating data from 6 gas stations in Blue Park and 5 gas stations in Green Spring. Use the chart below and determine if Salma's claim is supported! Table 10-3. Critical values of "t" significance level p= 0.05 significance level p= 0.01 TP 1 12.706 63.657 2 4.303 9.925 3 3.182 5.841 2.776 4.604 5 2.571 4.032 6 2.447 3.707 7 2.365 3.499 2.306 3.355 9 2.262 3.250 10 2.228 3.169 11 2.201 3.106 Copy the following into your text box, fill in the blanks, and answer the questions A. The degree of freedom is: B. The t-calculated value is: and the t-critical value is : C. Comparison: t-calc t-crit (answer with ()) D. Is the null hypothesis accepted or rejected? Explain how you determined that! Also, explain what it means to reject or accept the hypothesis! E. Explain if Salma's claim is supported based on your t-value…
Using the t-table, find the critical value for a left-tailed test with a significance level of 10% and a sample size of 25.
Suppose IQ scores were obtained for 20 randomly selected sets of couples. The 20 pairs of measurements yield x = 101.02, y = 100.75, r = 0.852, P-value = 0.000, and y = 5.93 + 0.94x, where x represents the IQ score of the husband. Find the best predicted value of y given that the husband has an lQ of 100? Use a significance level of 0.05. %3D Click the icon to view the critical values of the Pearson correlation coefficient r. The best predicted value of y is Critical values of the pearson correlation coefficient r (Round to two decimal places as needed.) Critical Values of the Pearson Correlation Coefficient r X = 0.05 a = 0.01 INOTE: To test Ho: p=0 n Jagainst H,: p+0, reject Ho if the absolute value of r is greater than the critical value in the table. 4 0.950 0.990 0.878 0.959 0.811 0.917 7 0.754 0.875 0.707 0.834 0.666 0.798 10 0.632 0.765 11 0.602 0.735 12 0.576 0.708 13 0.553 0.684 14 0.532 0.661 15 0.514 0.641 16 0.497 0.623 17 0.482 0.606 18 0.468 0.590 19 0.456 0.575 20 0.444…
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