# Lower Tail Test of Population Mean with Unknown Variance

The null hypothesis of the lower tail test of the population mean can be expressed as follows:

where μ0 is a hypothesized lower bound of the true population mean μ.

Let us define the test statistic t in terms of the sample mean, the sample size and the sample standard deviation s :

Then the null hypothesis of the lower tail test is to be rejected if t ≤−tα , where tα is the 100(1 α) percentile of the Student t distribution with n 1 degrees of freedom.

#### Problem

Suppose the manufacturer claims that the mean lifetime of a light bulb is more than 10,000 hours. In a sample of 30 light bulbs, it was found that they only last 9,900 hours on average. Assume the sample standard deviation is 125 hours. At .05 significance level, can we reject the claim by the manufacturer?

#### Solution

The null hypothesis is that μ 10000. We begin with computing the test statistic.

> xbar = 9900            # sample mean
> mu0 = 10000            # hypothesized value
> s = 125                # sample standard deviation
> n = 30                 # sample size
> t = (xbarmu0)/(s/sqrt(n))
> t                      # test statistic
[1] 4.3818

We then compute the critical value at .05 significance level.

> alpha = .05
> t.alpha = qt(1alpha, df=n1)
t.alpha               # critical value
[1] 1.6991

The test statistic -4.3818 is less than the critical value of -1.6991. Hence, at .05 significance level, we can reject the claim that mean lifetime of a light bulb is above 10,000 hours.

#### Alternative Solution

Instead of using the critical value, we apply the pt function to compute the lower tail p-value of the test statistic. As it turns out to be less than the .05 significance level, we reject the null hypothesis that μ 10000.

> pval = pt(t, df=n1)
> pval                   # lower tail pvalue
[1] 7.035e05