# Answer in Statistics and Probability for Darv #241078

Two types of rice varieties are being considered for yield and a comparison is needed. Thirty hectares were planted with the rice varieties exposed to fairly uniform growing conditions.

The results are tabulated below:

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Answer in Statistics and Probability for Darv #241078
Just from $13/Page variety A variety B average yield 80 sacks/hectare 35 sacks/hectare sample variance 5.9 12.1 At .05 significance level, can we conclude that variety A is the better type? “\begin{matrix}n & Variety\ A & Variety\ B \\n Average\ yield & 80\ sacks/hectare & 35\ sacks/hectare \\n Sample\ variance& 5.9 & 12.1 \\n\end{matrix}” The following null and alternative hypotheses need to be tested: “H_0:\mu_1\leq \mu_2” “H_1:\mu_1>\mu_2” This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. Based on the information provided, the significance level is “\alpha=0.05,” the degrees of freedom are computed as follows, assuming that the population variances are equal: “df=n_1+n_2-2=30+30-2=58.” The critical value for this right-tailed test “\alpha=0.05, df=58” degrees of freedom is “t_c=1.671553.” The rejection region for this right-tailed test is “R=\{t:t>1.671553\}.” Since it is assumed that the population variances are equal, the t-statistic is computed as follows: “t=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}” “=\dfrac{80-35}{\sqrt{\dfrac{(30-1)(5.9)^2+(30-1)(12.1)^2}{30+30-2}(\dfrac{1}{30}+\dfrac{1}{30})}}” “\approx18.309” Since it is observed that “t=18.309>1.671553=t_c,” it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value for right-tailed test “\alpha=0.05,” “df=58” degrees of freedom is “p\approx0,” and since “p=0<0.05=\alpha,” it is concluded that the null hypothesis is rejected. The degrees of freedom are computed as follows, assuming that the population variances are equal: “df=\dfrac{(\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2})^2}{\dfrac{(s_1^2/n_1)^2}{n_1-1}+\dfrac{(s_2^2/n_2)^2}{n_2-1}}” “=\dfrac{(\dfrac{(5.9)^2}{30}+\dfrac{(12.1)^2}{30})^2}{\dfrac{((5.9)^2/30)^2}{30-1}+\dfrac{((12.1)^2/30)^2}{30-1}}\approx42.052091” The critical value for this right-tailed test “\alpha=0.05, df=42.052091” degrees of freedom is “t_c=1.6819.” The rejection region for this right-tailed test is “R=\{t:t>1.6819\}.” Since it is assumed that the population variances are unequal, the t-statistic is computed as follows: “t=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}}” “=\dfrac{80-35}{\sqrt{\dfrac{(5.9)^2}{30}+\dfrac{(12.1)^2}{30}}}\approx18.309” Since it is observed that “t=18.309>1.671553=t_c,” it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value for right-tailed test “\alpha=0.05,” “df=42.052091” degrees of freedom is “p\approx0,” and since “p=0<0.05=\alpha,” it is concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean “\mu_1” is greater than “\mu_2,” at the “\alpha = 0.05” significance level. Therefore, there is enough evidence to conclude that variety A is the better type at the “\alpha = 0.05” significance level. # What Will You Get? We provide professional writing services to help you score straight A’s by submitting custom written assignments that mirror your guidelines. #### Premium Quality Get result-oriented writing and never worry about grades anymore. We follow the highest quality standards to make sure that you get perfect assignments. #### Experienced Writers Our writers have experience in dealing with papers of every educational level. 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