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To determine the sample mean, you need to add together all the values in the dataset and then divide by the total number of values. This method gives us an average, representing the central tendency of our data set. For our problem involving the life spans of Honolulu residents, we have 20 data points.
By calculating the sample mean, \[ \bar{x} = \frac{\sum x_i}{n} \]we ensure we understand the typical lifespan from our sample. In our case, summing up all the life spans and dividing by 20 yields a sample mean of approximately 71.4 years. This is a useful number showing the average life span of the sample group studied.

Standard deviation helps us understand how much individual data points deviate from our mean value. It's a measure of the spread of data in a sample, providing insight into variability. The formula for sample standard deviation is:\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \]Here, \(x_i\) represents each value in the dataset, and \(\bar{x}\) is the sample mean. By calculating this for our sample of Honolulu residents, we get a standard deviation of approximately 20.65 years. This suggests there is a significant range in the life spans within our sample, though most are around the mean.

The t-test allows us to determine if there is a statistically significant difference between the sample mean and a known population mean. For this exercise, we are testing whether the sample mean lifespan in Honolulu is different from the established mean of 77 years. The test statistic is calculated using:\[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]where \(\mu\) is the population mean, \(\bar{x}\) is the sample mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. Our calculated test statistic of approximately -1.22 informs whether the difference between our sample and population mean is due to random variation or another factor.

In hypothesis testing, we start by establishing two opposing statements: the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)). The null hypothesis assumes that there is no effect or difference, while the alternative proposes some kind of effect or difference. In our exercise:
- Null hypothesis (\(H_0\)): Mean lifespan is 77 years (\(\mu = 77\)).
- Alternative hypothesis (\(H_a\)): Mean lifespan is less than 77 years (\(\mu < 77\)).
These hypotheses provide a framework for our analysis, helping us decide if the evidence from our data supports a significant change or deviation from the established norm.

The critical value and significance level play crucial roles in hypothesis testing. The significance level (often denoted by \(\alpha\)) is the probability of rejecting the null hypothesis when it is true. For this exercise, we use a 5% significance level, which is quite typical.
The critical value corresponds to the threshold we compare the test statistic to and is derived from the t-distribution table based on our significance level and degrees of freedom (\(n-1\)). In our exercise, the critical value was approximately -1.729. Since the calculated test statistic of -1.22 does not fall into the critical region, we fail to reject the null hypothesis. This suggests that we do not have sufficient evidence to conclude that the population mean lifespan is less than 77 years.