Question

The decision in any hypothesis is to reject the null hypothesis OR to fail to reject the null hypothesis.

1. Under what conditions do you reject the null hypothesis in any hypothesis test?
2. Find an example of a hypothesis test of the mean in your homework this week and use it to explain how a conclusion is written if the null hypothesis is rejected?
3. Find an example of a hypothesis test of the mean in your homework this week and use it to explain how a conclusion is written if the null hypothesis is not rejected?

Answer 1

(a)

We reject the null hypothesis through any of the following two approaches.

• Critical value method- In this method we calculate test statistic and find critical value at given (or our considered) level of significance using standard table or any software. Then we reject null hypothesis if test statistic value lies outside the range of critical value.
• P-value approach- In this procedure we calculate test statistic and find corresponding p-value. If p-value is lesser than level of significance we reject null hypothesis.

(b)

Let us consider the following problem-

Data are collected in a clinical trial evaluating a new compound designed to improve wound healing in trauma patients. The new compound is compared against a placebo. After treatment for 5 days with the new compound or placebo, The extent of wound healing is measured and the data are shown in Table 7-6. Is there a difference in the extent of wound healing by Treatment? Run the appropriate test at a 5% level of significance.

Percent wound healing Treatment	0-25%	26-50%	51-75%	76-100%
New Compound	15	37	32	41
Placebo	36	45	34	10

Answer (approaching through critical value method)-

As we do not know population standard deviation (or variance) we have to perform two sample t-test to test equality of mean.

Suppose, the percent of wound healing using treatment is denoted random variable by X and that by placebo is denoted by random variable Y.

We know, class mark is average of lower and upper class limits.

Serial number (i)	Interval of percent of wound healing	Class mark (xi and yi)	Frequency for treatment (fx,i)	Frequency for placebo (fy,i)
1	0.5-25.5	13	15	36
2	25.5-50.5	38	37	45
3	50.5-75.5	63	32	34
4	75.5-100.5	88	41	10

We have to test for null hypothesis

against the alternative hypothesis

Our test statistic is given by

where

Level of significance

Degrees of freedom

In not equal type alternative hypothesis i.e. in two tailed test we reject null hypothesis if .

Here we see, .

So, we reject our null hypothesis.

Hence, based on the given data we can conclude that there is significant difference in percent of wound healing by treatment compared to the placebo used.

(c)

Let us consider the following problem-

A recent article claimed that the mean wedding cost in the US is $28,400. A random sample of 44 weddings from this year was taken from the western New York region. Test, at the 5% significance level, if the mean cost of a wedding in Western New York is different than the national average.

18600, 19100, 34600, 36900, 30900, 28700, 34700, 29200, 1500, 51700, 40400,

34000, 29100, 32700, 11600, 21700, 37000, 31400, 2200, 1300, 14700, 34000,

16700, 12100, 33900, 29100, 25100, 44800, 44700, 40700, 16600, 22000, 44700,

28100, 31100, 35100, 43800, 13200, 25600, 28700, 8000, 46400, 26900, 9700

Answer (approaching through p-value method)-

We do not know population standard deviation (or variance). So, we have to perform t-test.

We have to test for null hypothesis

against the alternative hypothesis

Our test statistic is given by

where

Degrees of freedom

P-value = 0.5846943    [Using R-code 'pt(-0.5506927,43)+1-pt(0.5506927,43)']

Level of significance

We reject null hypothesis if

Here, we see and so we cannot reject our null hypothesis.

So, our decision is as follows.

Since the p-value is larger than the significance level, we fail to reject the null hypothesis.

So, our conclusion is as follows.

At the 5% significance level, there is not sufficient evidence to conclude that the mean cost of weddings in western New York is different than the national mean cost of $28,400.