Question

HELP ME WITH THIS STATISTICS QUESTION, DONE IIN R PROGRAMMING.

A quality control engineer developed a control chart to maintain the weights of family size ketchup bottle (desired population median weight is 68 oz). Weights (in ounces) of 16 bottles taken at randomly selected times from the production process are as follows (data displayed in order of time of the selection): 68.2, 71.6, 69.3, 71.6 , 70.4, 65, 63.6, 64.7, 65.3, 64.2, 67.6, 68.6, 66.8,67.9, 66.7, 70.1 (a) (3 points) Do the data provide any evidence to conclude that the sample is not a random sample? Use a=0.05. Include the R function used, report the p-value and write your conclusion. (b) (3 points) Do Wilcoxon signed-rank test to see if the median score of the sampled population is less than 68 oz. Use alpha = 0.05. Include the R function used, p-value, and write your conclusion. (c) (2 points) What assumptions are necessary for the validity of the test in step (b)? Answer the above questions in the order displayed above.

Answer 1

(a). We shall use a sign test to test whether the sample is random with median 68 oz. If the sample is really random, then half of the values have to be above and half will be below median. We shall then check the numbers and for the significance using a binomial probability.

The R program and the output is given belwo:

x=c(68.2,71.6,69.3,71.6,70.4,65,63.6,64.7,65.3,64.2,67.6,68.6,66.8,67.9,66.7,70.1)
> m0=68
> sp=length(x[x>m0])
> sn=length(x[x<m0])
> n=sp+sn
> pv=pbinom(sp,n,0.5)
> sp
[1] 7
> pv
[1] 0.4018097
> n
[1] 16
>

The test statistic is S+=7, which is the number of observations above median out of n=16.

The p-value of the test is 0.4018>0.05, we fail to reject the null hypothesis. Hence, we conclude that the data is random.

(b). The r-code for testing the following Hypothesis:

$H_{0}:M_{D}=68$

$H_{1}:M_{D}<68$ where is the median.

Mo

The output from r is given belwo:

> wilcox.test(x,alter="less",mu=68)

Wilcoxon signed rank test with continuity correction

data: x
V = 56.5, p-value = 0.2847
alternative hypothesis: true location is less than 68

Warning message:
In wilcox.test.default(x, alter = "less", mu = 68) :
cannot compute exact p-value with ties
The p-value is 0.2847 for an one sided alternative. Since the p-value>0.05, we fail to reject the null hypothesis. Hence, we conclude that there is not enough evidence to claim that the median score of the sampled population is less than 68.

(c). The following assumptions are needed:

1. The data is independent

2. The distribution of values are approximately symmetrical.