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lukranit [14]
3 years ago
7

Rita is spending more time at home to study and practice math. Her efforts are finally paying off. On her first assessment she s

cored 58 points, then she scores 63 and 68 on her next two assessments. If her scores continued to increase at the same rate, on which assessments will she be scoring above 85? Select all that apply.
Mathematics
1 answer:
sammy [17]3 years ago
7 0
<span>On the first assessment Rita scored 58 points. For each next assessment the score increases by 5: 58 + 6 = 63, 63 + 5 = 68, 68 + 5 = 72, etc. So a1 = 58, d = 5. This is an example of the arythmetic sequence: a n = a 1 + ( n - 1 ) * d ; We have to find n such that: 58 + ( n - 1 ) * 5 > 85. ( n - 1 ) * 5 > 27; n - 1 > 5.4; n > 6.4. Finally: n = 7. Answer: Rita will be scoring above 85 on the 7th and each next assessment. </span>
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A random sample of 100 people from City A has an average IQ of 120 with a SD of 18. Independently of this, a random sample of 15
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Answer:

z=\frac{(120-116)-0}{\sqrt{\frac{18^2}{100}+\frac{15^2}{150}}}}=1.837

p_v =P(z>1.837)=1-P(Z  

Comparing the p value with a significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the average IQ on city A is signficantly higher than city B at 5% of singificance.  

Step-by-step explanation:

\bar X_{A}=120 represent the mean for sample 1

\bar X_{B}=116 represent the mean for sample 2

s_{A}=18 represent the sample standard deviation for 1  

s_{B}=15 represent the sample standard deviation for 2  

n_{A}=100 sample size for the group 2  

n_{B}=150 sample size for the group 2  

\alpha Significance level provided

z would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if residents of City A smarter on average, the system of hypothesis would be:  

Null hypothesis:\mu_{A}-\mu_{B}\leq 0  

Alternative hypothesis:\mu_{A} - \mu_{B}> 0  

We don't have the population standard deviation's, but the sample sizes are large enough we can apply a z test to compare means, and the statistic is given by:  

z=\frac{(\bar X_{A}-\bar X_{B})-\Delta}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}} (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:  

z=\frac{(120-116)-0}{\sqrt{\frac{18^2}{100}+\frac{15^2}{150}}}}=1.837

P value

Since is a one right tailed test the p value would be:  

p_v =P(z>1.837)=1-P(Z  

Comparing the p value with a significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the average IQ on city A is signficantly higher than city B at 5% of singificance.  

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Answer:

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Answer:

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