Question

Information about hand posture and forces generated by the fingers during manipulation of various daily objects is needed for designing high-tech hand prosthetic devices. The article "Grip Posture and Forces During Holding Cylindrical Objects with Circular Grips" (Ergonomies, 1996: 1163-1176) reported that for a sample of 11 females, the sample mean four-finger pinch strength (N) was 98.1 and the sample standard deviation was 14.2. For a sample of 15 males, the sample mean and sample standard deviation were 129.2 and 39.1, respectively.

a. A test carried out to see whether true average strengths for the two genders were different resulted in t=2.51 and P-value =.019. Does the appropriate test procedure described in this chapter yield this value of t and the stated P-value?

b. Is there substantial evidence for concluding that true average strength for males exceeds that for females by more than 25 N ? State and test the relevant hypotheses.

Short answer

(a) \(t = - 2.836,0.01 < P < 0.02\)

(b) There is not sufficient evidence to support the claim that the true average strength for males exceeds that for females by more than 25N.

Step-by-step-solution

Given in the output:

\(\begin{array}{l}{{\bar x}_1} = 98.1\\{s_1} = 14.2\\{{\bar x}_2} = 129.2\\{s_2} = 39.1\\{n_1} = 11\\{n_2} = 15\end{array}\)

Let us assume:

\(\alpha = 0.05\)

Step-by-step solution

To find the appropriate test procedure described in this chapter yield this value of t and the stated P-value 

(a)

Given claim: Differs

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain an equality.

\(\begin{array}{l}{H_0}:{\mu _1} = {\mu _2}\\{H_a}:{\mu _1} \ne {\mu _2}\end{array}\)

Determine the test statistic:

\(t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{98.1 - 129.2 - 0}}{{\sqrt {\frac{{14.{2^2}}}{{11}} + \frac{{39.{1^2}}}{{15}}} }} \approx - 2.836\)

Determine the degrees of freedom (rounded down to the nearest integer):

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{14.{2^2}}}{{11}} + \frac{{39.{1^2}}}{{15}}} \right)}^2}}}{{\frac{{{{\left( {14.{2^2}/11} \right)}^2}}}{{11 - 1}} + \frac{{{{\left( {39.{1^2}/15} \right)}^2}}}{{15 - 1}}}} \approx 18\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Student's T distribution in the appendix containing the t-value in the row d f=18 :

\(0.01 = 2 \times 0.005 < P < 2 \times 0.01 = 0.02\)

We then note that the given t-value is not correct

To State and test the relevant hypotheses.

(b)

Given claim: More than 25

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain an equality.

\(\begin{array}{l}{H_0}:{\mu _1} - {\mu _2} = - 25\\{H_a}:{\mu _1} - {\mu _2} < - 25\end{array}\)

Determine the test statistic:

\(t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{98.1 - 129.2 - ( - 25)}}{{\sqrt {\frac{{14.{2^2}}}{{11}} + \frac{{39.{1^2}}}{{15}}} }} \approx - 0.556\)

Determine the degrees of freedom (rounded down to the nearest integer):

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{14.{2^2}}}{{11}} + \frac{{39.{1^2}}}{{15}}} \right)}^2}}}{{\frac{{{{\left( {14.{2^2}/11} \right)}^2}}}{{11 - 1}} + \frac{{{{\left( {39.{1^2}/15} \right)}^2}}}{{15 - 1}}}} \approx 18\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Student's T distribution in the appendix containing the t-value in the row d f=18 :

\(P > 2 \times 0.10 = 0.20\)

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:

\(P > 0.05 \Rightarrow Fail to reject {H_0}\)

There is not sufficient evidence to support the claim that the true average strength for males exceeds that for females by more than 25 N.