A researcher believes that women today weigh less than in previous years. To investigate this belief, he randomly samples 41 adult women and records their weights. The scores have a mean of 111 lbs. and a standard deviation of 12.4. A local census taken several years ago shows the mean weight of adult women was 115 lbs.Refer to Exhibit 13-1. What do you conclude, using a = 0.011 tail?a) accept H1b) accept H0c) reject H0d) retain H0

Answer:Option B) Accept [tex]H_0[/tex]             Step-by-step explanation:We are given the following in the question:  Population mean, μ = 115 lbsSample mean, [tex]\bar{x}[/tex] =  111 lbsSample size, n = 41 Alpha, α = 0.01Sample standard deviation, s = 12.4 lbsFirst, we design the null and the alternate hypothesis[tex]H_{0}: \mu = 115\text{ lbs}\\H_A: \mu < 115\text{ lbs}[/tex] We use One-tailed t test to perform this hypothesis.Formula:[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex] Putting all the values, we have [tex]t_{stat} = \displaystyle\frac{111- 115}{\frac{12.4}{\sqrt{41}} } = -2.065[/tex] Now, [tex]t_{critical} \text{ at 0.01 level of significance, 40 degree of freedom } = -2.423[/tex] Since,                  [tex]t_{stat} > t_{critical}[/tex]We fail to reject the null hypothesis and accept the null hypothesis.