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Nesterboy
3 months ago
10

A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of awriting

machine) is at least 10 hours. A random sample of 18 pens is selected, the writing lifetime of each is determined,and a normal probability plot of the resulting data supports the use of a one-sample t test.a. What hypotheses should be tested if the investigators believe a priori that the design specification has been satisfied?b. What conclusion is appropriate if the hypotheses of part (a) are tested,t = —23, and a = .05?c. What conclusion is appropriate if the hypotheses of part (a) are tested, t = —1.8, and u = .01?d. What should be concluded if the hypotheses of part (a) are tested and t = —3.6 ?
Mathematics
1 answer:
tester [12.3K]3 months ago
5 0

Answer:

a) Null hypothesis:\mu \geq 10

Alternative hypothesis:\mu < 10

b) p_v =P(t_{17}

Given that the p-value is lower than the significance threshold in this situation, we have sufficient grounds to reject the null hypothesis.

c) p_v =P(t_{17}

In this case, since the p-value exceeds the significance threshold, we have adequate evidence to FAIL to reject the null hypothesis.

d) p_v =P(t_{17}

Here again, with the p-value being less than the significance level, we can reject the null hypothesis.

Step-by-step explanation:

1) Provided data and references

\bar X represents the average of the samples

s denotes the standard deviation of the samples

n=18 indicates the number of samples

\mu_o =10 is the value we are examining

\alpha defines the significance level for the test.

t represents the specific statistic of interest

p_v indicates the p-value relevant to the test (the variable of concern)

Define the null and alternative hypotheses.

To assess if the true mean is at least 10 hours, we must set up a hypothesis:

Part a

Null hypothesis:\mu \geq 10

Alternative hypothesis:\mu < 10

If we consider the sample size being less than 30 and the population deviation unknown, it’s more appropriate to use a t-test to compare the actual mean with the reference value, calculated as:

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

Part b

In this scenario t=-2.3, \alpha=0.05

Initially, we need to calculate the degrees of freedom df=n-1=18-1=17

Since this is a left-tailed test, the p-value is determined by:

p_v =P(t_{17}

In this instance, with the p-value being less than the significance level, we have sufficient evidence to reject the null hypothesis.

Part c

For this situation t=-1.8, \alpha=0.01

We need to find the degrees of freedom df=n-1=18-1=17

For the left-tailed test, the p-value is given by:

p_v =P(t_{17}

In this case, since the p-value is above the significance level, we have enough grounds to FAIL to reject the null hypothesis.

Part d

For this case t=-3.6, \alpha=0.05

Firstly, we find the degrees of freedom df=n-1=18-1=17

Since we are conducting a left-tailed test, the p-value is calculated as:

p_v =P(t_{17}

Here, with the p-value being lower than the significance threshold, we can reject the null hypothesis.

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