1 2 3 Find the following tvalues in the Critical Values of t

1)

2)

3)

Find the following t-values in the Critical Values of t table. (Enter your answers to three decimal places.) Organic chemists often purify organic compounds by a method known as fractional crystallization. One chemist prepared ten 4.85-g quantities of aniline and purified It to acetanilide. The following dry yields were recorded. 3.83 3.81 3.88 3.86 3.91 3.36 3.62 4.00 3.74 3.83 Estimate the mean grams of acetanilide that can be recovered from an initial amount of 4.85 g of aniline. Use a 95% confidence interval. (Round your answers to three decimal places.) (b) A right-tailed test with t = 3.30 and 19 df p-value

Solution

1b. A right tailed test with t = 3.30 and 19 df

Looking across the df 19 row of the t distribution table: 3.3 falls between 3.174 and 3.579, which are at 0.0025 and .001.

Since p is between .001 and .0025, p is less than .0025 and so is less than .005

answer: p < .005

1c. A right tailed test with a t = -1.12 and 24 df:

If you get negative t score, and the test is a right tailed test, then the p value is going to be high.

answer: p value > .2000

1d. A left tailed test with t = -6.77 and 7 df.

Looking across the 7 df row on the t distribution table, 6.77 is greater than 5.408 which is at .0005. so the p value is less than .0005.

answer: p value < .005

2. From the sample of 10, the mean of the sample is 3.784

the standard deviation of the sample is 0.17983

The t score to use for this 95% confidence interval is found on the distribution table under the .025 column and df of 9 row. This t is 2.262

confidence interval:

mean +/- t*sd/square root of n

3.784 +/- 2.262*.17983/square root of 10

3.784 +/- .1286

lower limit: 3.784 - .1286 = 3.655
Upper limit: 3.784 + .1286 = 3.913

answer: (3.655, 3.913)

3.
a. 2.132 (look under column .05, row df = 4)
b. 2.262 (look under column .025, df = 9)
c. 1.330 (look under column .10, df = 18)
d. 2.042 (look under column .025, df = 30)