Kelly runs a distance of 100 metres in a time of 10.52 seconds.
1 answer:
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Answer:
[895.05; 940.81]kWh
Step-by-step explanation:
Hello!
Be X: monthly electricity usage by a residential customer
X~N
σ²: 12100KWh²
To estimate the population mean using a 98% CI you have to use the following formula
[X[bar]± *
]
1-α: 0.98
α0.02
α/2: 0.01
1-α/2:0.99
n= 125
X[bar]= ∑X/n= 228565/125= 917.93
[917.93±2.326 * ]
[895.05; 940.81]
Using a 98% confidence level you'd expect that the interval [895.05; 940.81]kWh will include the average monthly electricity usage of residential customers.
I hope this helps!
Raw data:
Electric Usage
765
1139
714
687
1027
1109
749
799
911
631
975
717
1232
806
637
894
856
896
1272
1224
621
606
898
723
817
746
933
595
851
1027
770
685
750
1198
975
678
1050
886
826
1176
583
841
1188
692
733
791
584
1163
593
1234
603
1044
1233
1178
598
904
778
693
590
845
893
1028
975
788
1240
1253
854
1185
1164
741
1058
1053
795
1198
1240
1140
959
938
1008
1035
1085
1100
680
1006
977
1042
1252
943
1165
1014
912
791
612
935
864
953
667
1005
1063
1095
1086
810
1032
970
1099
1229
892
1074
579
754
1007
1116
583
763
1231
966
962
1132
738
1033
697
891
840
725
1031
Answer:
Step-by-step explanation:
Given :
Juan rolled a six-sided number cube 18 times.
The number two occurred four times.
To Find: Juan claimed the experimental probability of rolling a two was approximately 1/9. Why is Juan’s experimental probability incorrect?
Solution:
Total events = number of times cube rolled = 18
Favorable events = The number two occurred four times. = 4
So, Experimental probability =
=
=
Thus the experimental probability of rolling a two was
So, Juan’s experimental probability was incorrect.
Answer:
We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.
Step-by-step explanation:
Jenna claims that the average time of texting at her larger high school is greater than 94 minutes per day.
From here we can see that we have to perform a hypothesis test about a sample mean. The null and alternate hypothesis will be:
Null Hypothesis:
Alternate Hypothesis:
Jenna collected data from a sample of 32 students. So, sample size will be:
Sample Size = n = 32
Sample Mean = x = 96.5
Sample Standard Deviation = s = 6.3
We have to perform a hypothesis test, to test Jenna's claim. Since, the value of Population Standard Deviation is unknown and the value of Sample Standard Deviation is known, we will use One Sample t-test in this case.
The formula to calculate the test statistic is:
Using the values, we get:
The degrees of freedom will be:
df = n - 1 = 32 - 1 = 31
We have to convert the t-score 2.245 with 31 degrees of freedom to its equivalent p-value. From t-table this value comes out to be:
p-value = 0.0160
The significance level is:
Since, the p-value is lesser than the level of significance, we reject the Null Hypothesis.
Conclusion:
We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.
Answer:
There is enough statistical evidence to say that the means are different.
Step-by-step explanation:
<em>The question is incomplete:</em>
<em>"The two-sample t statistic for comparing the population means has value (±0.001)."</em>
<em />
The null and alternative hypothesis are:
The level of significance is assumed to be α=0.01.
The difference of the sample means is:
The standard deviation for the difference of the means is:
Then, the t-statistic is:
The degrees of freedom are:
With 40 degrees of freedom the t-critial for a significance of α=0.01 is t=±2.705.
As the t-statistic lies in the rejection region, the effect is significant and the null hypothesis is rejected.
There is enough statistical evidence to say that the means are different.