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baherus [9]
3 years ago
6

A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought i

n to the time repairs are completed. A random sample of 12 repair records showed the following repair times (in days): 9, 2, 5, 1, 5, 4, 7, 5, 11, 3, 7, 2. Click here for the Excel Data File (a) H0: μ ≤ 5 days versus H1: μ > 5 days. At α = .05, choose the right option. Reject H0 if tcalc > 1.796 Reject H0 if tcalc < 1.796 a bCalculate the Test statistic.
(c-1) The null hypothesis should be rejected.

TRUE
FALSE
(c-2) The average repair time is longer than 5 days.

TRUE
FALSE
(c-3) At \alpha = .05 is the goal being met?

Yes
No
Mathematics
1 answer:
diamong [38]3 years ago
8 0

Answer:

c-1) The null hypothesis should be rejected.

FALSE

c-2) The average repair time is longer than 5 days.

FALSE

c-3) At \alpha = .05 is the goal being met?

YES we fail to reject the null hypothesis

Step-by-step explanation:

Data given and notation  

\bar X=5.08 represent the mean for the sample  

s=2.999 represent the sample standard deviation for the sample  

n=12 sample size  

\mu_o =5 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is actually higher than 5, the system of hypothesis would be:  

Null hypothesis:\mu \leq 5  

Alternative hypothesis:\mu > 5  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{5.083-5}{\frac{2.999}{\sqrt{12}}}=0.0958    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=12-1=11  

Since is a one side test the p value would be:  

p_v =P(t_{(11)}>0.0958)=0.4627  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can conclude that the population mean is not significantly higher than 5.  

c-1) The null hypothesis should be rejected.

FALSE

c-2) The average repair time is longer than 5 days.

FALSE

c-3) At \alpha = .05 is the goal being met?

YES we fail to reject the null hypothesis

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B) Method 2 will increase the standard deviation of the students' scores.

Step-by-step explanation:

Given that an instructor in a college class recently gave an exam that was worth a total of 100 points.

The average score for his students was 43 and the standard deviation of the scores was 5 points.

And now he is considering two different strategies for rescaling the exam results of which:

Method 1 = Add 17 points to everyone's score.

Method 2 = Multiply everyone's score by 1.7 .

And we have to check what will be the impact of these methods on the standard deviation of the students' scores.

For this let us consider a simple example to understand this:

Firstly, Formula for calculating Standard Deviation =  \sqrt{\frac{\sum (X-Xbar)^{2}}{n-1}}

Suppose,

     X             X - Xbar       (X - Xbar)^{2}

     3              3 - 6 = -3         -3 * -3 = 9

     5              5 - 6 = -1          -1 * -1 = 1

     10            10 - 6 = 4          4 * 4 = 16

<em>Mean of above data, Xbar</em> = \frac{3+ 5+10}{3} = 6

<em>Standard Deviation of data </em>= \sqrt{\frac{26}{3-1} } = 3.6055

Now let us suppose that we multiply each value of above data with 2 so the new data will be:

     X                X - Xbar           (X - Xbar)^{2}

 3*2 = 6        6 - 12 = -6         -6 * -6 = 36

 5*2 = 10       10 - 12 = -2       -2 * -2 = 4

10*2 =20      20 - 12 = 8         8 * 8 = 64

<em>Mean of new data, Xbar </em>= \frac{6+ 10+20}{3} = 12

<em>Standard Deviation of new data</em> = \sqrt{\frac{104}{3-1} } = 7.2111

<em>Hence, we see that when we multiply any value to the data the standard deviation will increase and in other words it will multiplied by that value which value we multiplied with each data value i.e. when we multiply each data value with 2 the standard deviation also get multiplied by as </em>

  3.6055 * 2 = 7.2111

Therefore option B is correct that Method 2 will increase the standard deviation of the students' scores.

<em>And on the other hand Similarly by adding any constant to the data the Standard Deviation will remain same. Therefore Method 1 will have no impact on standard deviation of the students' scores.</em>

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