Invertase is an enzyme that may aid in spore germination of the fungus Colletotrichum graminicola. A

botanist incubated specimens of the fungal tissue in petri dishes and then assayed the tissue for invertase activity. The specific activity values for nine petri dishes incubated at 90% relative humidity for 24 hours are summarized as follows:¹⁵

(a) Assume that the data are a random sample from anormal population. Construct a 95% confidence

interval for the mean invertase activity under these experimental conditions.

(b) Interpret the confidence interval you found in part (a). That is, explain what the numbers in the interval

mean. (See Examples 6.3.4 and 6.3.5.)

(c) If you had the raw data, how could you check the condition that the data are from a normal population?

Example 6.3.4:

Bone Mineral Density Low bone mineral density often leads to hip fractures in the elderly. In an experiment to assess the effectiveness of hormone replacement therapy, researchers gave conjugated equine estrogen (CEE) to a sample of 94 women between the ages of 45 and 64.10 After taking the medication for 36 months, the bone mineral density was measured for each of the 94 women.The average density was 0.878 g/cm2, with a standard deviation of 0.126 g/cm².  he standard error of the mean is thus  . It is not clear that the distribution of bone mineral density is a normal distribution, but as we will see in Section 6.5, when the sample size is large, the condition of normality is not crucial. There were 94 observations, so there are 93 degrees of freedom.To find the multiplier

for a 95% confidence interval, we will use 100 degrees of freedom (since Table 4 doesn’t list 93 degrees of freedom); the multiplier is . A 95% confidence interval for  is

Thus, we are 95% confident that the mean hip bone mineral density of all women age 45 to 64 who take CEE for 36 months is between 0.852 g/cm2 and 0.904 g/cm².

Example 6.3.5:

Seeds per Fruit The number of seeds per fruit for the freshwater plant Vallisneria Americana varies considerably from one fruit to another.A researcher took a random sample of 12 fruit and found that the average number of seeds was 320, with a  tandard deviation of 125.11 The researcher expected the number of seeds to follow, at least approximately, a normal distribution. A normal probability plot of the data is shown in Figure 6.3.6. This supports the use of a normal distribution model for

these data.