Question

1. ​There are 6 Intro Statistics classes (say, groups) each with 15 students offered in the SP 2020 semester of a university. All the classes are taught by new graduate teaching assistants. The head of the department of Statistics wants to see if at least one class is significantly different from the overall instructional style. He decided to do an ANOVA F-test for the test marks. The ratio of variance between groups to variance within groups is 2.71. The P value of the test should be:         

1. 0.0034
2. 0.1315
3. 0.0255

2. The dynamic viscosity of the engine oil SAE 5W-40 at 0 Celsius follows a Uniform Distribution from 740 to 760 millipascal-seconds.

1. Take 16 samples of engine oil SAE 5W-40 and examine the dynamic viscosities at 0 Celsius: what should be the distribution of the mean dynamic viscosity? (You must define all the required parameters, and give the distribution in standard notations)   

2. For the 16 samples taken in step (d), what should be the probability that the sample mean dynamic viscosity is greater than 755 millipascal-seconds?   

Answer 1

ANSWER:

1)

​There are 6 Intro Statistics classes (say, groups) each with 15 students offered in the SP 2020 semester of a university

See this is an one way ANOVA. Here the effect of all 6 intro statistics courses are compared.

Here null hypothesis : all courses are equally effective

Alternative hypothesis: atleast one course is differently effective.

Let between group variance= sb and within group variance =sw and let the number of course be k=6 and total individuals studied=n =15*6=90.

Here test statistic is : Ft =

sb n k * SW k - 1

Under null hypothesis, Ft~F_k-1,n-k

Here sb/sw=2.71, k=6,n=90

So Ft ~ F_{​​​​​​5,84}

Observed Ft= 45.528

So P value is P(F_{​​​​​​5,84} > 45.528) = .0034 (using F table)

P value = 0.0034. (option -a)

2)

a)

We are given the underlying distribution here as:

X ~U(a = 740, b = 760

The mean and standard deviation here are computed as:

a + b E(X) = 740 + 760 2 750 2

SD(X) (b-a) V12 760 – 740 V12 5.7735

Therefore the distribution of the sample mean here is computed as:

X N N[// = 750, 7 T = 5.7735, V16

X Nu = 750,0 = 1.4434)

this is the required distribution of mean dynamic viscosity here.

b)

The probability here is computed as:

P(X 755)

Converting it to a standard normal variable, we have here:

P(Z > 755 - 750 1.4434

P(Z > 3.4641

Getting it from the standard normal tables, we have here:

P(Z > 3.4641) = 0.0003

Therefore 0.0003 is the required probability here.

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