Use the Stepwise Method 1.5 mg. dose (X,) 1.0 mg. dose Two groups of experimental rats were injected with tranquilizer at 1.0 mg. and 1.5mg. dose respectively. The time given in seconds that took them to fall asleep is hereby given. Use the t-test for independent samples at .01 to test the null hypothesis that the difference in dosage has no effect on the length of time it took them to fall asleep. (X,). 8.8 11.0 14.2 9.4 10.2 9.2 7.5 10.5 12.0 13.6 10.1 10.5 8.8 9.7 12.2 10.4 7.6 13.2 10.1 8.7

Use the Stepwise Method 1.5 mg. dose (X,) 1.0 mg. dose Two groups of experimental rats were injected with tranquilizer at 1.0 mg. and 1.5mg. dose respectively. The time given in seconds that took them to fall asleep is hereby given. Use the t-test for independent samples at .01 to test the null hypothesis that the difference in dosage has no effect on the length of time it took them to fall asleep. (X,). 8.8 11.0 14.2 9.4 10.2 9.2 7.5 10.5 12.0 13.6 10.1 10.5 8.8 9.7 12.2 10.4 7.6 13.2 10.1 8.7

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

4.27 please. I already had 4.26 answered. Thank you!
A hairdresser wanted to see if her clients bought more products at the end of their appointment if they were given lots of compliments. She decided to test this theory using an experiment. She randomly assigned 24 clients to receive either zero compliments, 5 compliments, or 10 compliments during their appointment. She recorded how many products they purchased after their appointment as her DV. She firstly hypothesised that no-compliment clients (M = 0.38) would purchase fewer products than clients who received 5 compliments (M = 1.50). She secondly hypothesised that 5-compliment clients would purchase fewer products than 10-compliment clients (M = 3.63). Thus, she conducted two comparisons. The initial ANOVA was significant. Using an alpha of .05, results for the follow-up tests confirmed her first hypothesis, with an obtained t value of 2.60, but did not confirm her second hypothesis, with an obtained t value of 1.20. She later realised that she did not use a Bonferroni-corrected…
refer to image blow
A treatment is administered to a sample selected from a population with a mean of μ = 80 and σ = 10. After the treatment, the sample mean M = 85. Based on this information, the effect size as measured by Cohen’s d is … Select one: a. 5.00 b. 2.00 c. .50 d. impossible to calculate without more information
(a) The figure to the right represents the normal curve with μ=266 days and σ=16 days. The area to the left of X=245 is 0.0947. Provide two interpretations of this area. Provide one interpretation of the area using the given values. Select the correct choice below and fill in the answer boxes to complete your choice. (Type integers or decimals.) A. The proportion of human pregnancies that last more than nothing days is nothing. B. The proportion of human pregnancies that last less than nothing days is nothing.
An engineer wants to know if producing metal bars using a new experimental treatment rather than the conventional treatment makes a difference in the tensile strength of the bars (the ability to resist tearing when pulled lengthwise). At α=0.10, answer parts (a) through (e). Assume the population variances are equal and the samples are random. If convenient, use technology to solve the problem. Treatment Tensile strengths (newtons per square millimeter) Experimental 449 354 450 360 433 388 400 Conventional 370 376 374 424 378 450 438 404 352 376 (a) Identify the claim and state H0 and Ha. The claim is "The new treatment ▼ makes a difference does not make a difference in the tensile strength of the bars." What are H0 and Ha? The null hypothesis, H0, is ▼ mu 1 equals mu 2μ1=μ2 mu 1 less than or equals mu 2μ1≤μ2 mu 1 greater than or equals mu 2μ1≥μ2 . The alternative hypothesis, Ha,…
The following table shows the hours studied and corresponding test grade earned by students on a recent test. Calculate the correlation coefficient, r, and determine whether r is statistically significant at the 0.01 level of significance. Round your answer to the nearest thousandth. Critical Values of the Pearson Correlation Coefficient Hours Studied and Test Grades Hours Studied 0 0.5 2 2.25 3.25 3.5 3.75 3.75 4 5 1 1 81 Test Grade 72 64 68 79 85 82 72 99 87 84 97
Given a normal distribution with u = 30 and o = 6, find the normal curve area to the left of x = 22. %3D Express your answer in four decimal places.
NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6. Step 1 of 2: Suppose a sample of 502 people is drawn. Of these people, 140 passed out at G forces greater than 6. Using the data, estimate the proportion of people who pass out at more than 6 Gs. Enter your answer as a fraction or a decimal number rounded to three decimal places.