91影视

Given:

\(\begin{array}{l}n = 65\\\bar x = 13.4\\{\sigma _{\bar x}} = 2.05\end{array}\)

Let us assume that we want to calculate the confidence bound with\(95\% \)confidence (other confidence levels work similarly).

\(c = 95\% = 0.95\)

Determine the t-value by looking in the row starting with degrees of freedom\(df = n - 1 = 65 - 1 = 64 > 60\)and in the column with\(1 - c = \)\(0.05\)in the table of the Student's T distribution:

\({t_\alpha } = 1.671\)

The margin of error is then:

\(E = {t_{\alpha /2}} \times \frac{s}{{\sqrt n }} = 1.671 \times 2.05 \approx 3.42555\)

The boundaries of the confidence interval then become:

\(\bar x + E = 13.4 + 3.42555 = 16.82555\)

Thus the \(95\% \) upper confidence bound is \(16.82555.\)

\(\begin{array}{l}{{\bar x}_1} = 13.4\\{{\bar x}_2} = 9.7\\{n_1} = 65\\{n_2} = 50\\{\sigma _{{{\bar x}_1}}} = 2.05 \Rightarrow {s_1} = {\sigma _{{{\bar x}_1}}}\sqrt n = 2.05\sqrt {65} \approx 16.5276\\{\sigma _{{{\bar x}_2}}} = 1.76 \Rightarrow {s_2} = {\sigma _{{{\bar x}_2}}}\sqrt n = 1.76\sqrt {50} \approx 12.4451\end{array}\)

Let us assume: \(\alpha = 0.05\)

Given claim: exceeds

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 the value mentioned in the claim.

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

Determine the test statistic:

\(t = \frac{{{{\bar x}_1} - {{\bar x}_2}}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{13.4 - 9.7}}{{\sqrt {\frac{{{{16.5276}^2}}}{{65}} + \frac{{{{12.4451}^2}}}{{50}}} }} \approx 1.369\)

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

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{{{16.5276}^2}}}{{65}} + \frac{{{{12.4451}^2}}}{{50}}} \right)}^2}}}{{\frac{{{{\left( {{{16.5276}^2}/65} \right)}^2}}}{{65 - 1}} + \frac{{{{\left( {{{12.4451}^2}/50} \right)}^2}}}{{50 - 1}}}} \approx 112 > 60\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The \({\rm{P}}\)-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 \(df = 60\) :

\(0.05 < P < 0.10\)

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

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

There is not sufficient evidence to support the claim that the true average time spent by blackbirds at the experimental location exceeds the true average time birds of this type spend at the natural location.

\({\bar x_1} = 9.7\)

\(\begin{array}{l}{{\bar x}_2} = 38.4\\{n_1} = 50\\{n_2} = 46\\{\sigma _{{{\bar x}_1}}} = 1.76 \Rightarrow {s_1} = {\sigma _{{{\bar x}_1}}}\sqrt n = 1.76\sqrt {50} \approx 12.4451\\{\sigma _{{{\bar x}_2}}} = 5.06 \Rightarrow {s_2} = {\sigma _{{{\bar x}_2}}}\sqrt n = 5.06\sqrt {46} \approx 34.3186\end{array}\)

Let us assume that we want to calculate the confidence interval with $95 \%$ confidence (other confidence levels work similarly).

\(c = 95\% = 0.95\)

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

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{{{12.4451}^2}}}{{50}} + \frac{{{{34.3186}^2}}}{{46}}} \right)}^2}}}{{\frac{{{{\left( {{{12.4451}^2}/50} \right)}^2}}}{{50 - 1}} + \frac{{{{\left( {{{34.3186}^2}/46} \right)}^2}}}{{46 - 1}}}} \approx 55 > 50\)

Determine the t-value by looking in the row starting with degrees of freedom\(df = 50\)and in the column with\(1 - c/2 = 0.025\)in the Student's distribution table in the appendix:

\({t_{\alpha /2}} = 2.009\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot \sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} = 2.009 \cdot \sqrt {\frac{{{{12.4451}^2}}}{{50}} + \frac{{{{34.3186}^2}}}{{46}}} \approx 10.7629\)

The endpoints of the confidence interval for \({\mu _1} - {\mu _2}\) are: \(\begin{array}{l}\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E = (1.5083 - 1.5875) - 10.7629 = - 0.0792 - 10.7629 = - 10.8421\\\left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E = (1.5083 - 1.5875) + 10.7629 = - 0.0792 + 10.7629 = 10.6837\end{array}\)