Question

A certain enzyme in the liver of fish is considered an indicator of trace amounts of a dangerous pollutant that is difficult to detect by chemical methods. Enzyme activities of less than 50 units per gram of liver (fresh weight) are taken to indicate the presence of the pollutant. A random sample was taken from a local stream and the enzyme concentrations were as follows 48 43 51 42 50 42 44 45 56 49 44 47 50 49 38 46 38 52 32 56 Compute a 95% and 99% confident interval for the mean enzyme activity for the fish population in this stream. (You will have to include the conf.level 0.99 in the R code for the t.test. 2. Do a hypothesis test on this data. 3. Are enzyme activities depressed in this population? Now let's look at the same test with paired data.

R code and answer questions please

Answer 1

Answer:

1).

R code

# single sample t

x <- c(48,43,51,42,50,42,44,45,56,49,44,47,50,49,38,46,38,52,32,56)

t.test(x,mu=50,alternative = "two", conf.level=.95)

t.test(x,mu=50,alternative = "two", conf.level=.99)

R output

One Sample t-test

data: x

t = -2.8946, df = 19, p-value = 0.009287

alternative hypothesis: true mean is not equal to 50

95 percent confidence interval:

43.28003 48.91997

sample estimates:

mean of x

     46.1

> t.test(x,mu=50,alternative = "two", conf.level=.99)

        One Sample t-test

data: x

t = -2.8946, df = 19, p-value = 0.009287

alternative hypothesis: true mean is not equal to 50

99 percent confidence interval:

42.24541 49.95459

sample estimates:

mean of x

     46.1

2).

Lower tail test.

R code

t.test(x,mu=50,alternative = "l", conf.level=.95)

        One Sample t-test

data: x

t = -2.8946, df = 19, p-value = 0.004644

alternative hypothesis: true mean is less than 50

95 percent confidence interval:

     -Inf 48.42969

sample estimates:

mean of x

     46.1

Calculated t= -2.8946, P=0.0046 which is < 0.05 level of significance. Ho is rejected.

We conclude that enzyme activities is less than 50 units per gram of liver.