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the lengths of time (in years) it took a random sample of 32 former smo…

Question

the lengths of time (in years) it took a random sample of 32 former smokers to quit smoking permanently are listed. assume the population standard deviation is 4.3 years. at \\( \alpha = 0.01 \\), is there enough evidence to reject the claim that the mean time it takes smokers to quit smoking permanently is 13 years? complete parts (a) through (e).
10.7 22.7 17.7 14.4 19.6 21.7 11.8 9.9
18.4 10.7 9.5 9.6 13.3 22.3 15.4 21.4
8.1 14.2 12.6 14.4 11.2 16.3 8.7 19.3
20.8 7.2 18.9 22.4 7.1 9.4 8.3 7.8
(a) identify the claim and state the null hypothesis and alternative hypothesis.
a. \\( h _ { 0 } : \mu = 13 \\) (claim) \\( h _ { a } : \mu \
eq 13 \\)
b. \\( h _ { 0 } : \mu > 13 \\) \\( h _ { a } : \mu \leq 13 \\) (claim)
c. \\( h _ { 0 } : \mu \geq 13 \\) (claim) \\( h _ { a } : \mu < 13 \\)
d. \\( h _ { 0 } : \mu \leq 13 \\) (claim) \\( h _ { a } : \mu > 13 \\)
e. \\( h _ { 0 } : \mu > 13 \\) (claim) \\( h _ { a } : \mu \leq 13 \\)
f. \\( h _ { 0 } : \mu \
eq 13 \\) (claim) \\( h _ { a } : \mu = 13 \\)
(b) identify the standardized test statistic. use technology.
\\( z = \\) (round to two decimal places as needed.)

Explanation:

Step1: Calculate the sample mean

First, sum up all the data values.

$$ LATEXBLOCK0 $$

The sample size \(n = 32\). The sample mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}=\frac{476.8}{32}=14.9\)

Step2: Calculate the z - statistic

The formula for the z - statistic in a one - sample z - test is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\), where \(\mu = 13\) (population mean), \(\sigma=4.3\) (population standard deviation), and \(n = 32\) (sample size), \(\bar{x}=14.9\) (sample mean)

$$ LATEXBLOCK1 $$

Answer:

\(z = 2.50\)