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AlexFokin [52]
4 years ago
9

A sample of 84 full electric cars was randomly drawn from a population with an unknown mean range μ (miles per charge) and a sta

ndard deviation σ of 25 (miles per charge). The cars in the sample have an average range of 125 (miles per charge). (a)[9] At the level of significance α = 0.04, test H0: μ ≥ 130 versus H1: μ < 130. Sketch the test. (b)[5] Sketch & find the p‐value. Would you accept the null if: α = 0.01; α = 0.05; and α = 0.10.
Mathematics
1 answer:
forsale [732]4 years ago
3 0

Answer:

(a) The test statistic value is -1.83.

(b) We accept the null hypothesis at α = 0.01.

Step-by-step explanation:

The information provided is:

 n=84\\\bar x=125\\\sigma=25\\\alpha =0.04

The hypothesis is defined as follows:

<em>H₀</em>: <em>μ</em> ≥ 130 versus <em>Hₐ</em>: <em>μ</em> < 130.  

Since the population standard deviation is provided, we will use a z-test.

Compute the test statistic as follows:

 z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}=\frac{125-130}{25/\sqrt{84}}=-1.83

The test statistic value is -1.83.

(b)

Compute the p-value of the test as follows:

p-value = P (Z < -1.83)

= 0.034

* Use a z-table.

We reject a hypothesis if the p-value of a statistic is lower than the level of significance α.

• p-value = 0.034 < α = 0.04. Reject the null hypothesis.

• p-value = 0.034 > α = 0.01. Fail to reject the null hypothesis.

• p-value = 0.034 < α = 0.05. Reject the null hypothesis.

• p-value = 0.034 < α = 0.10. Reject the null hypothesis.

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Answer:

C. \frac{1}{18}

Step-by-step explanation:

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Solution:

Sample space for sum of cards when two cards are drawn at random is \{(1,1),(1,2),(1,3)......(6,6)\}

total number of possible cases =36

Sample space when sum of cards is 8 is \{(3,5),(5,3),(6,2),(2,6),(4,4)\}

Total number of possible cases =5

Sample space when one of the cards is 5 is \{(5,3),(3,5)\}

Total number of possible cases =2

Let A be the event that sum of cards is 8

p(\text{A}) =\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}

p(\text{A})=\frac{5}{36}

Let B be the event when one of the two cards is 5

probability than one of two cards is 5 when sum of cards is 8

p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}

p(\frac{\text{B}}{\text{A}})=\frac{2}{5}

Now,

probability that sum of cards 8 is and one of cards is 5

p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})

p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}

p(\text{A and B})=\frac{1}{18}

if sum of cards is 8 then probability that one of the cards is 8 is \frac{1}{18}, option C is correct.

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