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Veseljchak [2.6K]

2 years ago

10

A survey of 32 different gas stations in Texas found the average price of gasoline to be $3.19 a gallon with a sample standard d

eviation of $0.096. Let μ be the true average price of gasoline price in Texas. The survey company claimed that gasoline in Texas was significantly lower than the national average $3.26 a gallon. The alernative hypothesis to test the claim is H 1 : μ <$3.26. Calculate the test statistic.

1 answer:

Leni [432]2 years ago

3 0

Answer:

The value of test statistic is -4.1247

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = $3.26 a gallon

Sample mean, $\bar{x}$ = $3.19 a gallon

Sample size, n = 32

Sample standard deviation, σ = $0.096

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 3.26\text{ dollars a gallon}\\H_A: \mu < 3.26\text{ dollars a gallon}$

We use one-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{3.19 - 3.26}{\frac{0.096}{\sqrt{32}} } = -4.1247$

Thus, the value of test statistic is -4.1247

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Answer:

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Step-by-step explanation:

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The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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This means that $p = 0.3$

A sample of 25 dishwasher detergent customers was taken.

This means that $n = 25$

a. What is the probability that 10 or fewer customers choose the leading brand?

This is:

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

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$P(X = 1) = C_{25,1}.(0.3)^{1}.(0.7)^{24} = 0.0014$

$P(X = 2) = C_{25,2}.(0.3)^{2}.(0.7)^{23} = 0.0074$

$P(X = 3) = C_{25,3}.(0.3)^{3}.(0.7)^{22} = 0.0243$

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$P(X = 5) = C_{25,5}.(0.3)^{5}.(0.7)^{20} = 0.1030$

$P(X = 6) = C_{25,6}.(0.3)^{6}.(0.7)^{19} = 0.1472$

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$P(X = 8) = C_{25,8}.(0.3)^{8}.(0.7)^{17} = 0.1651$

$P(X = 9) = C_{25,9}.(0.3)^{9}.(0.7)^{16} = 0.1336$

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0.9021 = 90.21% probability that 10 or fewer customers choose the leading brand

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