Tyra is laying down square carpet pieces in her photography studio. Each square carpet piece is 1 yard by 1 yard. If Tyra's phot
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Answer:
<em>Mean of the sample = 27.83</em>
<em> The variance of the the sample = 106.96</em>
<em> </em><em>Standard deviation of the sample = 10.34</em>
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given random sample of six employees
x 26 32 29 16 45 19
mean of the sample
Mean of the given data = 27.83
<u>Step(ii):-</u>
<u>Given data</u>
x : 26 32 29 16 45 19
x - x⁻ : -1.83 4.17 1.17 -11.83 17.17 -8.83
(x - x⁻)² : 3.3489 17.3889 1.3689 139.9489 294.80 77.9689
∑ (x-x⁻)² = 534.8245
Given sample size 'n' =6
The variance of given data
S² = ∑(x-x⁻)² / n-1
The variance of the given sample = 106.9649
<u> Step(iii):-</u>
Standard deviation of the given data
Standard deviation of the sample = 10.3423
Answer:
We conclude that the new procedure will not decrease the population mean amount of time required to produce the part.
Step-by-step explanation:
We are given that a large manufacturing plant has analyzed the amount of time required to produce an electrical part and determined that the times follow a normal distribution with mean time μ = 45 hours.
A random sample of 25 parts will be selected and the average amount of time required to produce them will be determined. The sample mean amount of time is = 43.118 hours with the sample standard deviation s = 5.5 hours
<em>Let </em><em> = population mean amount of time required to produce an electrical part using new procedure</em>
SO, <u>Null Hypothesis</u>, :
45 hours {means that the new procedure will remain same or increase the population mean amount of time required to produce the part}
<u>Alternate Hypothesis,</u> :
< 45 hours {means that the new procedure will decrease the population mean amount of time required to produce the part}
The test statistics that will be used here is <u>One-sample t test statistics </u>because we don't know about the population standard deviation;
T.S. = ~
where, = sample mean amount of time = 43.118 hours
s = sample standard deviation = 5.5 hours
n = sample of parts = 25
So, <u><em>test statistics</em></u> = ~
= -1.711
<em>Now at 0.025 significance level, the t table gives critical value of -2.064 at 24 degree of freedom for left-tailed test. Since our test statistics is more than the critical value of t so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.</em>
Therefore, we conclude that the new procedure will remain same or increase the population mean amount of time required to produce the part.