PROBLEM 1:A company claims that the mean weight per banana it ships is 160 grams withstandard deviation of 18 grams. Data generated from a sample of 49 bananas randomlyselected from a shipment indicated a mean weight of 153.3 grams per banana. Do the 49selected bananas have lower weight than the rest? Test the claim at 0.05 level of significant.Formulate the hypothesis1. Ho:2. HA:3. Given:5. Solve for z:n =a =4. Test statistics:6. z =7. Solve for confidence level9.8. Confidence level:-11.3.

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Answered: PROBLEM 1: A company claims that the…

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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For a population with a mean of ? = 80 and a standard deviation of ? = 12, a score of X = 77 corresponds to z = −0.50. True or false?
For a population with mean of 50 and standard deviation of 8, what is the raw score that corresponds to z=-1.50? а. 42 b. 52 C. 70 d. 38 е. 46
Please answer items D to H only. I really need help
4. The weights of oranges in a harvest are normally distributed, with a mean weight of 150 grams and standard deviation of 10 grams. Calculate the z-score for an orange weighing a. 160 grams b. 135 grams Round your answers to one decimal place.
3.2/2
For a population with mean of 50 and standard deviation of 8, what is the z- score that corresponds to X=44? а. -2 b. 1.5 С. 1 d. -0.5 e. -0.75
Step 1 of 3. State the null and alternative hypotheses for the test. Step 2 of 3. Compute the value of the test statistic. Round your answer to three decimal places Step 3 of 3. Draw a conclusion and interpret the decision.
There are many situations in which we want to compare means from populations having standard deviations that are equal. This method applies even if the standard deviations are known to be only approximately equal. Consider a report regarding average incidence of fox rabies in two regions. For region I, n1 = 16, x1 ≈ 4.73, and s1 = 2.84 and for region II, n2 = 15, x2 ≈ 3.93, and s2 = 2.45. The two sample standard deviations are sufficiently close that we can assume ?1 = ?2. Use the method of pooled standard deviation to consider the report, testing if there is a difference in population mean average incidence of rabies at the 5% level of significance. (Round your answer to three decimal places.) t =
You suspect that people living in urban areas produce on average more garbage (X)than those living in rural areas (Y). To verify this opinion, you draw two samples and collect information on the amount of garbage produced in a month (measured in kilos). The first sample, made of 8 subjects living in an urban area, provides a sample mean equal to 40.5 and a sample standard deviation equal to 5.6; in the second sample, made of 10 subjects living in rural areas, the sample mean is 35.6 and the sample standard deviation is 7.5. It is assumed that the two populations are normal with the same unknown variance.Test the null hypothesis that the two population means are equal against a one-sided alternative according to which people living in urban areas produce on average more garbage than those living in rural areas. Use a 1% significance level.
A standardized test is given to a sixth-grade class. Historically the mean score has been 151 with a standard deviation of 21. The superintendent believes that the standard deviation of performance may have recently decreased. She randomly sampled 23 students and found a mean of 161 with a standard deviation of 18.2898. Is there evidence that the standard deviation has decreased at the α=0.05 level? Step 1 of 5 : State of the hypotheses in terms of the standard deviation. Round the standard deviation to four decimal places when necessary Step 2 of 5 : Determine the critical value(s) of the test statistic. If the test is two-tailed, separate the values with a comma. Round your answer to three decimal places. Step 3 of 5 : Determine the value of the test statistic. Round your answer to three decimal places. Step 4 of 5 : Make the decision. Step 5 of 5 : What is the conclusion?
Solve question 9, Part A, Part B and Part C only.
3.1. Suppose that the height of the women is a normal variable with a mean of 1.62 meters (m) and a standard deviation of 24 centimeters. Calculate the percentage of women who measures (a) more than 1.70 m. (b) between 1.66 and 1.58 m. (c) less than 1.50 m. d) between 1.60 and 1.63 m.

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