GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.24 mg of mercury. A sample of 25 bulbs shows a mean of 3.29

Question

GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.24 mg of mercury. A sample of 25 bulbs shows a mean of 3.29 mg of mercury.
(a) State the hypotheses for a right-tailed test, using GreenBeam’s claim as the null hypothesis about the mean.

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Vivian 1 week 2021-10-09T18:54:31+00:00 1 Answer 0

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    2021-10-09T18:55:47+00:00

    Answer:

    Null hypothesis:\mu \leq 3.24  

    Alternative hypothesis:\mu > 3.24  

    Step-by-step explanation:

    1) Previous concepts  and data given

    A confidence interval is “a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval”.

    The margin of error is the range of values below and above the sample statistic in a confidence interval.

    Normal distribution, is a “probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean”.

    A right tailed test (sometimes called an upper test) is when the alternative hypothesis statement contains a greater than (>) symbol.

    \bar X=3.29 represent the sample mean  

    s represent the sample standard deviation

    n represent the sample selected

    \alpha significance level  

    State the null and alternative hypotheses.  

    We need to conduct a hypothesis in order to check if the mean for fluorescent bulbs is no more than 3.24 mg of mercury, the system of hypothesis would be:  

    Null hypothesis:\mu \leq 3.24  

    Alternative hypothesis:\mu > 3.24  

    IIf we know the population deviation we can apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

    z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

    z-test: “Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value”.  

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27:3+15-4x7+3-1=? ( )