Mathematics, 31.07.2020 04:01 ssollers

A group of students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are listed below. Use a 0.01 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one minute?

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Answer from: aroyalstudent

Complete Question

The complete question is shown on the first uploaded image

Yes the students are reasonably good at estimating one minute

Option A is correct

The test statistics is t = 0.354

Step-by-step explanation:

From the question we are told that

The set of data is

68, 82 , 38 , 62 , 41, 25 , 57 , 64, 67, 47, 61, 71, 91, 87, 64

The population mean is $\mu = 60$

The level of significance is given as $\alpha = 0.01$

The critical value for this level of significance obtained from the normal distribution table is

$Z_{\alpha } = 2.33$

The null hypothesis is

$H_o : \mu = 60 \ seconds$

The alternative hypothesis is

$Ha : \mu \ne 60 \ seconds$

Generally the sample mean is mathematically represented as

$\= x = \frac{\sum x_i }{n }$

where n = 15

$\= x = \frac{ 68+ 82 + 38 + 62 + 41+ 25 + 57 + 64+67+ 47+ 61+ 71+ 91+ 87+ 64}{15}$

$\= x = 61.67$

The standard deviation is mathematically represented as

$\sigma =\sqrt{ \frac{ \sum (x_i - \= x )^2}{n} }$

substituting values

$\sigma =\sqrt{ \frac{ \sum (68-61.67)^2 + ( 82-61.67 )^2 + (38 -61.67)^2 + 62-61.67)^2 + (41-61.67)^2 +(25-61.67)^2 +( 57+61.67)^2 }{15} }$ $\sqrt{ \frac{ ( \cdot \cdot + ( 64-61.67)^2 + (67-61.67)^2 +(47-61.67)^2 + (61-61.67)^2+ (71-61.67)^2 + (91 -61.67)^2+( 87-61.67)^2 + (64-61.67)^2}{15} }$ => $\sigma = 18.23$

The test statistics is evaluated as

$t = \frac{\= x - \mu }{ \frac{\sigma }{\sqrt{n} } }$

substituting values

$t = \frac{61.67 -60 }{ \frac{18.23 }{\sqrt{ 15} } }$

t = 0.354

Now comparing the statistics and the critical value of the level of significance we see that the the test statistics is less than the critical value

Hence the we fail to reject the null hypothesis which mean that these times are from a population with a mean equal to 60 seconds

So we can state that yes the students are reasonably good at estimating one minute given that the sample mean is not far from the population mean

A group of students estimated the length of one minute without reference to a watch or clock, and th

Answer from: Quest

option a.

step-by-step explanation:

if you look closely at the graph, we can notice that the function is not defined at two different points.

if you look at functions b and c, you will notice that both functions are not defined at just one point. the function b is not defined at x= -5/2 and function c is not defined at x=-5. both options are discarded, given that our function should not be defined at two fiferent points.

now, if you look closely at the graph you will notice that the function sketched is not defined at x=5 and x=-2.

Which of the following rational functions is graphed below ?

Answer from: Quest

kekdkdkccc

step-by-step explanation:

kekdkckcldlx

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