Medical practitioners are increasingly utilising the Zoom platform. Not only for online meetings but also for consultations with patients. The weekly Zoom access time (in hours) at two hospitals:Health Care I Health Care II needs to be investigated.Given: The weekly access time of medical practitioners at both hospitals is normally distributed with a standard deviation of σ=6 hours Consider the weekly access time at Health Care IIThe Superintendent wants to investigate the average weekly access time (in hours) at Health Care II based on 2 different research questions. Arandom sample of the weekly access times of n = 20 medical practitioners revealed an average ofi = 26.61 hours.Let: µ = mean weekly access time of the medical practitioners.Research Question 1: Is the mean weekly access time less than 28 hours?Given: The value of the test statistic isz = -1.04Complete the following:Identify the correct null hypothesis:О Но : и + 28О Но : и 3 28О Но : и 3D 26.61O Ho :i < 28O Ho :> 28O Ho :i+ 26.61O Ho : i = 28О Но : и < 28O Ho :i> 26.61О Но : и + 26.61O Ho :š < 26.61O Ho : x # 28О Но : и> 28O Ho :š = 26.61О Но : и > 26.61O Ho : µ < 26.61The p-value is:Round your answer to 4 decimal places.Research Question 2: Does the mean weekly access time differ from 22 hours?Given: The 90% confidence interval for u is:(24.4 , 28.8) Complete the following:According to the observed confidence interval, the null hypothesis isand there is sufficient evidence to claim that theaverage access timefrom 22 hours, at 0.1 level of significance.According to the observed confidence interval, we know that:True False90% of all samples should have sample means between 24.4 and 28.8.For every 100 random samples drawn, we expect that 90 of the confidence intervals willcontain the population mean.90% of all medical practitioners at Health Care Il have access times between 24.4 and 28.8hours.For every 100 random samples drawn, we expect that 90 of the confidence intervals willcontain the sample mean.We are 90% confident that u is between 24.4 and 28.8.The probability that u is between 24.4 and 28.8 is 0.9.O OO o O

Answered: The Superintendent wants to investigate…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

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Question

Medical practitioners are increasingly utilising the Zoom platform. Not only for online meetings but also for consultations with patients. The weekly Zoom access time (in hours) at two hospitals:

Health Care I
Health Care II

needs to be investigated.

Given: The weekly access time of medical practitioners at both hospitals is normally distributed with a standard deviation of σ=6 hours

Consider the weekly access time at Health Care II
The Superintendent wants to investigate the average weekly access time (in hours) at Health Care II based on 2 different research questions. A
random sample of the weekly access times of n = 20 medical practitioners revealed an average ofi = 26.61 hours.
Let: µ = mean weekly access time of the medical practitioners.
Research Question 1: Is the mean weekly access time less than 28 hours?
Given: The value of the test statistic is
z = -1.04
Complete the following:
Identify the correct null hypothesis:
О Но : и + 28
О Но : и 3 28
О Но : и 3D 26.61
O Ho :i < 28
O Ho :> 28
O Ho :i+ 26.61
O Ho : i = 28
О Но : и < 28
O Ho :i> 26.61
О Но : и + 26.61
O Ho :š < 26.61
O Ho : x # 28
О Но : и> 28
O Ho :š = 26.61
О Но : и > 26.61
O Ho : µ < 26.61
The p-value is:
Round your answer to 4 decimal places.
Research Question 2: Does the mean weekly access time differ from 22 hours?
Given: The 90% confidence interval for u is:
(24.4 , 28.8)

Complete the following:
According to the observed confidence interval, the null hypothesis is
and there is sufficient evidence to claim that the
average access time
from 22 hours, at 0.1 level of significance.
According to the observed confidence interval, we know that:
True False
90% of all samples should have sample means between 24.4 and 28.8.
For every 100 random samples drawn, we expect that 90 of the confidence intervals will
contain the population mean.
90% of all medical practitioners at Health Care Il have access times between 24.4 and 28.8
hours.
For every 100 random samples drawn, we expect that 90 of the confidence intervals will
contain the sample mean.
We are 90% confident that u is between 24.4 and 28.8.
The probability that u is between 24.4 and 28.8 is 0.9.
O O
O o O

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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