Question

A researcher would like to determine whether a new tax on legalized marijuana has had any effect on people’s purchasing behavior. During the year before the tax was imposed, medical marijuana stores reported selling an average of µ = 410 grams of marijuana per day with a σ = 60 grams. The distribution of daily sales was approximately normal. For a sample of n = 9 days following the new tax, the researcher found an average of M = 386 grams of marijuana per day for the same stores. Use a two-tailed test with α = .05
a. Is the sample mean sufficient to conclude that there was a significant change in marijuana purchases after the new tax? Use a two-tailed test with α = .05.
b. If the population standard deviation was σ = 30, is the result sufficient to conclude that there is a significant difference in sales of marijuana?

Answer 1

Answer:

a) $z=\frac{386-410}{\frac{60}{\sqrt{9}}}=-1.2$    

$p_v =2*P(z$  

If we compare the p value and the significance level given $\alpha=0.05$ 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 conclude that the true mean is not different from 410 at 5% of signficance.  

b) $z=\frac{386-410}{\frac{30}{\sqrt{9}}}=-2.4$  

Since is a two-sided test the p value would be:  

$p_v =2*P(z$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is significantly different from 410 at 5% of signficance.  

Step-by-step explanation:

Part a

Data given and notation  

$\bar X=386$ represent the sample mean    

$\sigma=60$ represent the population standard deviation

$n=9$ sample size  

$\mu_o =410$ represent the value that we want to test  

$\alpha=0.05$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean differs from 410, the system of hypothesis would be:  

Null hypothesis:$\mu =410$  

Alternative hypothesis:$\mu \neq 410$  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

$z=\frac{386-410}{\frac{60}{\sqrt{9}}}=-1.2$  

P-value  

Since is a two-sided test the p value would be:  

$p_v =2*P(z$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ 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 conclude that the true mean is not different from 410 at 5% of signficance.  

Part b

$z=\frac{386-410}{\frac{30}{\sqrt{9}}}=-2.4$  

P-value  

Since is a two-sided test the p value would be:  

$p_v =2*P(z$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is significantly different from 410 at 5% of signficance.