A Web Usage Snapshot indicated a monthly average of 34 Internet visits of a particular website peruser from home. A random sample of 30 Internet users yielded a sample mean of 35.9 visits with astandard deviation of 4.3.At the 0.05 level of significance, can it be concluded that this differs from the national average?Assume the population is normally distributed.Express your answers correct to 3 decimal places.a) What test should be performed (z or t)? Type the corresponding letter.b) Find the critical value(s). If there are two values (2 –tail test), type the negative value first andseparate them with a comma.c) Calculate the test statisticd) Make a decision. Type R if you choose to "Reject the null hypothesis"; Type F if you choose to"Fail to reject the null hypothesis".e) Choose the correct conclusion (Type 1 or 2):1. At 5% level of significance, there is not enough evidence to support the claim that the mean isdifferent from the national average.2. At 5% level of significance, there is enough evidence to support the claim that the mean isdifferent from the national average.

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A Web Usage Snapshot indicated a monthly average of 34 Internet visits of a particular website per user from home. A random sample of 30 Internet users yielded a sample mean of 35.9 visits with a standard deviation of 4.3. At the 0.05 level of significance, can it be concluded that this differs from the national average? Assume the population is normally distributed.

A Web Usage Snapshot indicated a monthly average of 34 Internet visits of a particular website per user from home. A random sample of 30 Internet users yielded a sample mean of 35.9 visits with a standard deviation of 4.3. At the 0.05 level of significance, can it be concluded that this differs from the national average? Assume the population is normally distributed.

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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