a nutritionist claims that the mean tuna consumption by a person is 3.9 pounds per year. a sample of 60…

a nutritionist claims that the mean tuna consumption by a person is 3.9 pounds per year. a sample of 60 people shows that the mean tuna consumption by a person is 3.7 pounds per year. assume the population standard deviation is 1.19 pounds. at \\( \\alpha = 0.07 \\), can you reject the claim?\n(a) identify the null hypothesis and alternative hypothesis.\na. \\( h _ { 0 } : \\mu = 3.9 \\)\n\\( h _ { a } : \\mu \\neq 3.9 \\)\nb. \\( h _ { 0 } : \\mu \\leq 3.9 \\)\n\\( h _ { a } : \\mu > 3.9 \\)\nc. \\( h _ { 0 } : \\mu > 3.7 \\)\n\\( h _ { a } : \\mu \\leq 3.7 \\)\nd. \\( h _ { 0 } : \\mu \\leq 3.7 \\)\n\\( h _ { a } : \\mu > 3.7 \\)\ne. \\( h _ { 0 } : \\mu > 3.9 \\)\n\\( h _ { a } : \\mu \\leq 3.9 \\)\nf. \\( h _ { 0 } : \\mu \\neq 3.7 \\)\n\\( h _ { a } : \\mu = 3.7 \\)\n(b) identify the standardized test statistic.\n\\( z = \\square \\) (round to two decimal places as needed.)

a nutritionist claims that the mean tuna consumption by a person is 3.9 pounds per year. a sample of 60 people shows that the mean tuna consumption by a person is 3.7 pounds per year. assume the population standard deviation is 1.19 pounds. at \\( \\alpha = 0.07 \\), can you reject the claim?\n(a) identify the null hypothesis and alternative hypothesis.\na. \\( h _ { 0 } : \\mu = 3.9 \\)\n\\( h _ { a } : \\mu \\neq 3.9 \\)\nb. \\( h _ { 0 } : \\mu \\leq 3.9 \\)\n\\( h _ { a } : \\mu > 3.9 \\)\nc. \\( h _ { 0 } : \\mu > 3.7 \\)\n\\( h _ { a } : \\mu \\leq 3.7 \\)\nd. \\( h _ { 0 } : \\mu \\leq 3.7 \\)\n\\( h _ { a } : \\mu > 3.7 \\)\ne. \\( h _ { 0 } : \\mu > 3.9 \\)\n\\( h _ { a } : \\mu \\leq 3.9 \\)\nf. \\( h _ { 0 } : \\mu \\neq 3.7 \\)\n\\( h _ { a } : \\mu = 3.7 \\)\n(b) identify the standardized test statistic.\n\\( z = \\square \\) (round to two decimal places as needed.)

Answer

Explanation:

Step1: Recall the formula for 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 (\bar{x}) is the sample mean, (\mu) is the population mean under the null hypothesis, (\sigma) is the population standard deviation, and (n) is the sample size.

Step2: Substitute the given values into the formula

We are given that (\bar{x} = 3.7), (\mu=3.9), (\sigma = 1.19), and (n = 60). [ \begin{align*} z&=\frac{3.7 - 3.9}{\frac{1.19}{\sqrt{60}}}\ &=\frac{- 0.2}{\frac{1.19}{7.746}}\ &=\frac{-0.2}{0.1549}\ &\approx - 1.29 \end{align*} ]

Answer:

(z=-1.29)