A skateboarding club randomly surveyed 200 people who visited the club's website about their favorite skateboard brand. The data
collected is shown in the table.
1 answer:
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<span>Let's analyze Hannah's work, step-by-step, to see if she made any mistakes.
</span>In Step 1, Hannah wrote
<span> as the sum of two separate derivatives </span>
<span>using the </span><span>sum rule.
</span>
This step is perfectly fine.
In Step 2,
was kept as it is, and 
was rewritten as 
using the constant rule.Indeed, according to the constant rule, the derivative of a constant number is equal to zero.
This step is perfectly fine.
In Step 3,
was rewritten as 
supposedly using the constant multiple rule.
The problem is that according to the constant multiple rule,
should be rewritten as 
and not as 
.
<span>Therefore, Hannah made a mistake in this step.</span>
</span>In Step 1, Hannah wrote
</span>
This step is perfectly fine.
In Step 2,
This step is perfectly fine.
In Step 3,
The problem is that according to the constant multiple rule,
<span>Therefore, Hannah made a mistake in this step.</span>
Answer:
(a) Null Hypothesis, :
= $1,150
Alternate Hypothesis, :
$1,150
(b) The test statistic is 3.571.
(c) We conclude that the mean of all account balances is significantly different from $1,150.
Step-by-step explanation:
We are given that a sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.
We have test the hypothesis to determine whether the mean of all account balances is significantly different from $1,150.
<u><em>Let </em></u><u><em> = mean of all account balances</em></u>
(a)So, Null Hypothesis, :
= $1,150 {means that the mean of all account balances is equal to $1,150}
Alternate Hypothesis, :
$1,150 {means that the mean of all account balances is significantly different from $1,150}
The test statistics that will be used here is <u>One-sample t test statistics</u> as we don't know about population standard deviation;
T.S. = ~
where, = sample average balance = $1,200
s = sample standard deviation = $126
n = sample of account balances = 81
(b) So, <u>test statistics</u> = ~
= 3.571
The value of the sample test statistics is 3.571.
(c) <u>Now, P-value of the test statistics is given by the following formula;</u>
P-value = P( > 3.571) = <u>Less than 0.05%</u>
Because in the t table the highest critical value for t at 80 degree of freedom is given between 3.460 and 3.373 at 0.05% level.
Now, since P-value is less than the level of significance as 5% > 0.05%, so we sufficient evidence to reject our null hypothesis.
Therefore, we conclude that the mean of all account balances is significantly different from $1,150.