Question

The process that produces Sonora Bars (a type of candy) is intended to produce bars with a mean weight of 56 gm. The process standard deviation is known to be 0.77 gm. A random sample of 49 candy bars yields a mean weight of 55.82 gm. Which are the hypotheses to test whether the mean is different than it is supposed to be? Assume that candy bar weights are normally distributed.
Select one:
a. H0: μ ≤ 56, HA: μ > 56
b. H0: μ ≥ 56, HA: μ < 56
c. H0: μ = 56, HA: μ ≠ 56
d. H0: μ < 56, HA: μ ≥ 56

Answer 1

Answer:

c. H0: μ = 56, HA: μ ≠ 56

Step-by-step explanation:

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

1) Data given and notation      

$\bar X=55.82$ represent the mean weight for the Canby Bars

$\sigma=0.77$ represent the standard deviation for the population    

$n=49$ sample size      

$\mu_o =56$ represent the value that we want to test    

$\alpha$ 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 determine if the mean is different than it is supposed to be (56), the system of hypothesis would be:    

Null hypothesis:$\mu = 56$      

Alternative hypothesis:$\mu \neq 56$      

We know the population deviation, so for this case 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:      

$z=\frac{55.82-56}{\frac{0.77}{\sqrt{49}}}=-1.636$      

Calculate the P-value      

Since is a one-side lower test the p value would be:      

$p_v =P(z<-1.636)=0.05092$  

Conclusion      

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean weight not differs significantly from the standard 56.