Fruit flies, like almost all other living organisms, have built-in circadian rhythms that keep time even in the absence of external stimuli. Several genes have been shown to be involved in internal timekeeping, including per (period) and tim (timeless). Mutations in these two genes, and in other genes, disrupt timekeeping abilities. Interestingly, these genes have also been shown to be involved in other time-related behavior, such as the frequency of wingbeats in male courtship behaviors. Individuals that carry particular mutations of per and tim have been shown to copulate for longer than individuals that have neither mutation. But do these two mutations affect copulation time in similar ways? The following table summarizes some data on the duration of copulation for flies that carry either the tim mutation or the per mutation (Beaver and Giebultowicz 2004).

Mutation Mean copulation duration (min) Standard deviation of copulation duration Sample size, n
per 17.5 3.37 14
tim 19.9 2.47 17

Assume that the copulation time data are normally distributed. Do tim and per mutations lead to different mean copulation durations? Determine the absolute value of the test statistic for the appropriate test.

Answer:

The decision Rule is  

Reject the null hypothesis

The conclusion is  

There is sufficient evidence to states that  tim and per's mutations lead to different mean copulation duration.

The  test statistics is  [tex]z =-2.2187[/tex]

Step-by-step explanation:

From the question we are told that

    The Mutation Mean copulation duration is for  per [tex]\= x_1 = 17.5[/tex]

     The Mutation Mean copulation duration is for tim [tex]\= x_2 = 19.9[/tex]

     The Mutation Standard deviation copulation duration for is  per [tex]s _1 = 3.37[/tex]

      The Mutation Standard deviation copulation duration for is   tim [tex]s _2 = 2.47[/tex]

       The sample size for per is  [tex]n_1 = 14[/tex]

       The sample size for tim is  [tex]n_2 = 17[/tex]

The null hypothesis  is  [tex]H_o : \mu_1 = \mu_2[/tex]

The alternative  hypothesis  is  [tex]H_a : \mu_1 \ne \mu_2[/tex]

Generally the test statistics is mathematically represented as

      [tex]z = \frac{\=x_1 - \= x_2 }{ \sqrt{\frac{s_1^2 }{n_1} + \frac{s_2^2 }{n_2} } }[/tex]

=>   [tex]z = \frac{17.5 - 19.9 }{ \sqrt{\frac{3.37^2 }{14} + \frac{2.47^2 }{17} } }[/tex]

=>   [tex]z =-2.2187[/tex]

From the z table  the area under the normal curve to the left corresponding to    -2.2187  is

    [tex]P(Z < -2.2187) = 0.013254[/tex]

Generally the p-value is mathematically represented as

      [tex]p-value = 2 * P(Z <-2.2187 )[/tex]

=>    [tex]p-value = 2 * 0.013254[/tex]

=>    [tex]p-value = 0.02651[/tex]

Let assume that the level of significance is  [tex]\alpha = 0.05[/tex]

So

 From the value  obtained we see that [tex]p-value < \alpha[/tex] hence  

The decision Rule is  

Reject the null hypothesis

The conclusion is  

There is sufficient evidence to states that  tim and per's mutations lead to different mean copulation duration.

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