Question

Nitrites are often added to meat products as preservatives. In a study of the effect of these chemicals on bacteria, the rate of uptake of a radio-labeled amino acid was measured for a number of cultures of bacteria, some growing in a medium to which nitrites had been added. Here are the summary statistics from this study. Group n x s Nitrite 30 7880 1115 Control 30 8112 1250.

Carry out a test of the research hypothesis that nitrites decrease amino acid uptake at the 2% significance level.

Answer 1

Answer:

H0:$\mu_{N}\geq \mu_{C}$

H1:$\mu_{N} < \mu_{C}$

$t=\frac{7880-8112}{\sqrt{\frac{1115^2}{30}+\frac{1250^2}{30}}}}=-0.759$  

$p_v =P(t_{(58)}<-0.759)=0.225$

So the p value is a very High value and using any significance level given $\alpha=0.02$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the mean for the nitrite group is significantly lower than the mean for th control group .

Step-by-step explanation:

1) Data given and notation

$\bar X_{N}=7880$ represent the mean for the sample Nitrite

$\bar X_{C}=8112$ represent the mean for the sample control

$s_{N}=115$ represent the sample standard deviation for the sample Nitrite

$s_{C}=1250$ represent the sample standard deviation for the sample control

$n_{N}=30$ sample size for the group nitrite

$n_{C}=30$ sample size for the group control

t would represent the statistic (variable of interest)

$\alpha=0.02$ represent the significance level

Confidence =1-0.02=0.98 or 98%

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the nitrites decrease amino acid uptake , the system of hypothesis would be:

H0:$\mu_{N}\geq \mu_{C}$

H1:$\mu_{N} < \mu_{C}$

If we analyze the size for the samples both are equal to 30, but we don't know the population deviations, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{N}-\bar X_{C}}{\sqrt{\frac{s^2_{N}}{n_{N}}+\frac{s^2_{C}}{n_{C}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

We can replace in formula (1) like this:

$t=\frac{7880-8112}{\sqrt{\frac{1115^2}{30}+\frac{1250^2}{30}}}=-0.759$  

4) Statistical decision

The first step is calculate the degrees of freedom, on this case:

$df=n_{N}+n_{C}-2=30+30-2=58$

Since is a left tailed test the p value would be:

$p_v =P(t_{(58)}<-0.759)=0.225$

So the p value is a very High value and using any significance level given $\alpha=0.02$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the mean for the nitrite group is significantly lower than the mean for th control group .