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Karo-lina-s [1.5K]
3 years ago
11

Jo.

Mathematics
1 answer:
AleksAgata [21]3 years ago
3 0
Oh okie so sorry sorry about the delay in
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mariarad [96]

Answer:

p_v =P(z>2.087) =0.0184

So with the p value obtained and using the significance level given \alpha=0.01 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the mean for the control group it's not significantly higher than the mean of the treatment group by 1 point.  

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

\sigma=2.5

And the statistic is given by this formula:

z=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sigma\sqrt{\frac{1}{n_1}}+\frac{1}{n_2}}

Where z follows a normal standard distribution

The system of hypothesis on this case are:

Null hypothesis: \mu_c \leq \mu_t+1

Alternative hypothesis: \mu_c >\mu_t+1

Or equivalently:

Null hypothesis: \mu_c - \mu_t \leq 1

Alternative hypothesis: \mu_c-\mu_t>1

Our notation on this case :

n_c =45 represent the sample size for group control

n_t =45 represent the sample size for group  treatment

\bar X_c =5.2 represent the sample mean for the group control

\bar X_t =3.1 represent the sample mean for the group treatment

And now we can calculate the statistic:

z=\frac{(5.2-3.1)-(1)}{2.5\sqrt{\frac{1}{45}}+\frac{1}{45}}=2.087  

And now we can calculate the p value using the alternative hypothesis:

p_v =P(z>2.087) =0.0184

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